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Sum in base-12 javascript [duplicate]

I would like to convert numbers between different bases, such as hexadecimal and decimal.
Example: How do you convert hexadecimal 8F to decimal?

The API
To convert to a number from a hex string:
parseInt(string, radix)
string: Required. The string to be parsed
radix: Optional. A number (from 2 to 36) that represents the numeral system to be used
To convert from a number to a hex string:
NumberObject.toString(radix)
radix: Optional. Specifies the base radix you would like the number displayed as.
Example radix values:
2 - The number will show as a binary value
8 - The number will show as an octal value
16 - The number will show as an hexadecimal value
Example Usage
Integer value to hex:
var i = 10;
console.log( i.toString(16) );
Hex string to integer value:
var h = "a";
console.log( parseInt(h, 16) );
Integer value to decimal:
var d = 16;
console.log( d.toString(10) );

Update 2023-01-05: Now supports large numbers and floats via https://github.com/ryansmith94/baseroo
I came to this post needing to convert from base 10 to 62 and vice-versa. Whilst the solutions here are great, parseInt and toString only support base 2 to 36. So if anyone finds themselves in a similar position to me needing base 2 to 62, I've pasted my solution below.
function convertBase(value, from_base, to_base) {
var range = '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ+/'.split('');
var from_range = range.slice(0, from_base);
var to_range = range.slice(0, to_base);
var dec_value = value.split('').reverse().reduce(function (carry, digit, index) {
if (from_range.indexOf(digit) === -1) throw new Error('Invalid digit `'+digit+'` for base '+from_base+'.');
return carry += from_range.indexOf(digit) * (Math.pow(from_base, index));
}, 0);
var new_value = '';
while (dec_value > 0) {
new_value = to_range[dec_value % to_base] + new_value;
dec_value = (dec_value - (dec_value % to_base)) / to_base;
}
return new_value || '0';
}

You may try the following code, which also supports arbitrary precision numbers (larger than 2^53).
function convertBase(str, fromBase, toBase) {
const DIGITS = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ+/";
const add = (x, y, base) => {
let z = [];
const n = Math.max(x.length, y.length);
let carry = 0;
let i = 0;
while (i < n || carry) {
const xi = i < x.length ? x[i] : 0;
const yi = i < y.length ? y[i] : 0;
const zi = carry + xi + yi;
z.push(zi % base);
carry = Math.floor(zi / base);
i++;
}
return z;
}
const multiplyByNumber = (num, x, base) => {
if (num < 0) return null;
if (num == 0) return [];
let result = [];
let power = x;
while (true) {
num & 1 && (result = add(result, power, base));
num = num >> 1;
if (num === 0) break;
power = add(power, power, base);
}
return result;
}
const parseToDigitsArray = (str, base) => {
const digits = str.split('');
let arr = [];
for (let i = digits.length - 1; i >= 0; i--) {
const n = DIGITS.indexOf(digits[i])
if (n == -1) return null;
arr.push(n);
}
return arr;
}
const digits = parseToDigitsArray(str, fromBase);
if (digits === null) return null;
let outArray = [];
let power = [1];
for (let i = 0; i < digits.length; i++) {
digits[i] && (outArray = add(outArray, multiplyByNumber(digits[i], power, toBase), toBase));
power = multiplyByNumber(fromBase, power, toBase);
}
let out = '';
for (let i = outArray.length - 1; i >= 0; i--)
out += DIGITS[outArray[i]];
return out;
}
Usage:
console.log(convertBase("5a2a9c826c75045be9ba8fbffc80c6f25a2a9c826c75045be9ba8fbffc80c6f2",16,64));
// Returns: 5EGD89ItghrWrGfL/O0NL9qaFO2r7k4m+CWzX/YwcrO
console.log(convertBase("5EGD89ItghrWrGfL/O0NL9qaFO2r7k4m+CWzX/YwcrO",64,16));
// Returns: 5a2a9c826c75045be9ba8fbffc80c6f25a2a9c826c75045be9ba8fbffc80c6f2
The basic code has been found here, I have a bit improved it to also support up to base 64.

The following diagram might help. Note that to convert from base 16 to base 2 you need to convert first to base 10 and then base 2.

This function generate decimal number to any base from 2 to 36.(as like javascript)But you can Increase the base more than 36 just by adding new character in keys[] like lowercase["a","b"]
function toBase(num, radix = 10) { // only i64 numbers
var keys = ['🤢', '😃', 2, 3, 4, 5, 6, 7, 8, 9, "A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z"];
if (!(radix >= 2 && radix <= keys.length)) throw new RangeError("toBase() radix argument must be between 2 and " + keys.length)
if (num < 0) var isNegative = true
if (isNaN(num = Math.abs(+num))) return NaN
let output = [];
do {
let index = num % radix;
output.unshift(keys[index]);
num = Math.trunc(num / radix);
} while (num != 0);
if (isNegative) output.unshift('-')
return output.join("");
}
console.log(toBase("100",2))

Specify the radix you want to use as a parameter.
NOTE: This only works to convert from bases 2-36 to decimal and little values.
parseInt(string, radix)
parseInt("80", 10) // results in 80
parseInt("80", 16) // results in 128
// etc
About "little", parseInt("6f", 32) is fine (= 207), but any other little bigger will be also 207, 6f1, 6f11, ...

Well, I made a function that could translate from base 10 to any base. (This depends on how many strings you have in the array A, if it's more than that + 10 it'll run out of symbols), and I almost cried when I found out you could to it in less than 10 characters with that...
Add a bookmark and as URL insert this... I've done it the long but personal way. At least, mine can use a base which is higher than 36. You can add more symbols yourself, but if you want, I can make it for you...
var X = prompt("Choose your number");
var Y = prompt("Choose your base");
var Z = [];
var M = -1;
var A = ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z"];
var B = function() {
for (i = X; 0 < i; i = Math.floor(i / Y)) {
if(i % Y >= 10) {
Z.push(A[i % Y - 10]);
} else {
Z.push(i % Y);
}
M = M + 1;
}
for (j = M; j >= 0; j--) {
document.write(Z[j]);
}
};
B(); // Call function

Usually I use this function to convert from different bases.
For example, it returns "11111111" for both cases:
convertBase("ff", 16, 2) or convertBase(0xFF, 16, 2)
var convertBase = function(val, base1, base2) {
if (typeof(val) == "number") {
return parseInt(String(val)).toString(base2);
} else {
return parseInt(val.toString(), base1).toString(base2)
};
}

I've written a function to convert a JavaScript string from one base to another base, with the original base and the new base specified as parameters.
function convertFromBaseToBase(str, fromBase, toBase){
var num = parseInt(str, fromBase);
return num.toString(toBase);
}
alert(convertFromBaseToBase(10, 2, 10));

This function converts a number from base 10, to an arbitrary base:
function to_base(base, num) {
const largest_power = ~~(Math.log(num) / Math.log(base));
const result = [];
for (let pow = largest_power; pow >= 0; pow--) {
const digit = ~~(num / base ** pow);
num -= digit * base ** pow;
result.push(digit);
}
return result;
}
to_base(2, 13) // [1, 1, 0, 1]
to_base(10, 458) // [4, 5, 8]
to_base(32, 1024) // [1, 0, 0]
to_base(32, 1023) // [31, 31]
which can be useful if you want to use weird bases along with weird char-set

Using the parseInt function:
var noInBase10 = parseInt('8F',16);

Check the complete JS code to convert into different base
/**
* Convert From/To Binary/Decimal/Hexadecimal in JavaScript
* https://gist.github.com/shamshul2007/
* Copyright 2012-2015, Shamshul <shamshul2007#gmail.com>
* Licensed under The MIT License
* http://www.opensource.org/licenses/mit-license
*/
(function(){
var ConvertBase = function (num) {
return {
from : function (baseFrom) {
return {
to : function (baseTo) {
return parseInt(num, baseFrom).toString(baseTo);
}
};
}
};
};
// binary to decimal
ConvertBase.bin2dec = function (num) {
return ConvertBase(num).from(2).to(10);
};
// binary to hexadecimal
ConvertBase.bin2hex = function (num) {
return ConvertBase(num).from(2).to(16);
};
// decimal to binary
ConvertBase.dec2bin = function (num) {
return ConvertBase(num).from(10).to(2);
};
// decimal to hexadecimal
ConvertBase.dec2hex = function (num) {
return ConvertBase(num).from(10).to(16);
};
// hexadecimal to binary
ConvertBase.hex2bin = function (num) {
return ConvertBase(num).from(16).to(2);
};
// hexadecimal to decimal
ConvertBase.hex2dec = function (num) {
return ConvertBase(num).from(16).to(10);
};
//Octal to Decimal
ConvertBase.oct2dec = function (num) {
return ConvertBase(num).from(8).to(10);
};
//Decimal to Octal
ConvertBase.dec2oct = function (num) {
return ConvertBase(num).from(10).to(8);
};
this.ConvertBase = ConvertBase;
})(this);
/*
* Usage example:
* ConvertBase.bin2dec('1111'); // '15'
* ConvertBase.dec2hex('82'); // '52'
* ConvertBase.hex2bin('e2'); // '11100010'
* ConvertBase.dec2bin('153'); // '10011001'
* ConvertBase.hex2dec('1FE4ED63D55FA51E'); //'2298222722903156000'
* ConvertBase.oct2dec('777'); //'511'
* ConvertBase.dec2oct('551'); //'1047'
*/

Try the following code, optimised from Slavik Meltser's post, implements BASE n conversions with all the radix combinations between Base2 and Base256. This code accepts three sorts of arguments to define source and destination number systems:
by number system radix (e.g. 8)
by number system convention name (e.g. 'Bitcoin')
by giving custom numerals as an argument (e.g. ['0123456789ABCDEF'])
You'll see that some number system numerals have been hard-coded inside the class and will be used as default when you pass the radix as an argument. (.e.g. 64) When no hard-coded numeral exists (e.g. 16), default numerals are assigned for all radixes ranging between Base2 and Base256, which becomes very clear in the self-test further below.
function BASE() {
/**
* BASE n converter doing all the radix combinations between Base2 and Base256
* #param {String} str input number
* #param {Number|String|Array} fromBase input number system radix (Number, e.g. 64), convention name (String, e.g. 'Bitcoin') or range (Array)
* #param {Number|String|Array} toBASE output number system radix (Number), convention name (String) or range (Array e.g. ['0123456789'])
* #return {String} output number
*/
this.convert = function (str, fromBase, toBASE)
{
if(typeof(fromBase)=='object') { this.fromSymbols = fromBase[0] } else this.fromSymbols = this.getsymbols(fromBase);
if(typeof(toBASE) =='object') { this.toSymbols = toBASE[0] } else this.toSymbols = this.getsymbols(toBASE);
fromBase = this.fromSymbols.length; toBASE = this.toSymbols.length;
// PARSE INPUT DIGITS ARRAY
for(var _a = [0], str = str.split(''); str.length > 0 && _a[_a.push(this.fromSymbols.indexOf(str.pop())) - 1] >= 0;);
var _d = _a.shift() + _a[_a.length-1]>=0 ? _a : null; if (_d === null) return null;
// BASE CONVERSION
for (var _n = 0,_a = [],_p = [1]; _n < _d.length; _n++) { _a = add(_a, mul(_d[_n], _p, toBASE), toBASE); _p = mul(fromBase, _p, toBASE) }
// PARSE OUTPUT DIGITS ARRAY
for (var _n = _a.length - 1, _o = ''; _n >= 0; _o += this.toSymbols[_a[_n--]]);
return _o.length==0?this.toSymbols[0]:_o;
}
this.symbols = {
32:function(){return this["base32hex"]},
36:["[0-9][A-Z]"],
45:function(){return this["qr-alnum"]},
58:function(){return this["Bitcoin"]},
64:["[A-Z][a-z][0-9]+/"],
85:function(){return this["RFC 1924"]},
91:["[A-Z][a-z][0-9]!#$%&()*+,./:;<=>?#[]^_`{|}~\""],
94:["[!-~]"],
"geohash": ["[0-9][b-h]jkmn[p-z]"], // base 32
"RFC 4648": ["[A-Z][2-7]"], // base 32
"base32hex": ["[0-9][A-V]"], // base 32
"qr-alnum":["[0-9][A-Z] $%*+-./:"], // base 45
"Bitcoin": ["[1-9][A-H]JKLMN[P-Z][a-k][m-z]"], // base 58
"RFC 1924": ["[0-9][A-Z][a-z]!#$%&()*+-;<=>?#^_`{|}~"] // base 85
}
this.getsymbols = function(index) {
if(typeof(this.symbols[index])=="undefined") this.symbols[index] = index<95?this.rng(index<64?"[0-9][A-Z][a-z]+":"[A-Z][a-z][0-9][!-/][:-#][[-`][{-~]").substring(0,index):this.rng("[\x00-\xff]").substring(256-index,256);
if(typeof(this.symbols[index])=="function") this.symbols[index] = this.symbols[index](); // process references
if(typeof(this.symbols[index])=="object") this.symbols[index] = this.rng(this.symbols[index][0]); // process range_replace
return this.symbols[index];
}
this.rng = function(_s) {
var _a = _s.match(/\[.-.\]/); if(_a==null) return _s; else { _a=[_a[0].charCodeAt(1),_a[0].charCodeAt(3)];
return this.rng(_s.replace(RegExp("\\[(\\x"+("0"+_a[0].toString(16)).slice(-2)+"-\\x"+_a[1].toString(16)+")\\]","g")
,String.fromCharCode(..." ".repeat(_a[1]-_a[0]+1).split("").map((_e,_i)=>_i+_a[0])) )) }
}
this.selftest = function() {
var _a={}; for(var _o in this.symbols) _a[_o] = this.getsymbols(_o).length; // built-in symbols
for(_o=2;_o<=95;_o++) _a[_o] = this.getsymbols(_o).length; _a[256]=256; // symbol range 2-95 + 256 (96-255 is similar)
var _s = "",_a = Object.keys(_a).sort(function(a,b){return _a[a]-_a[b]}); // sort merged list
for(var _i in _a) { // iterate number systems
_o = {fromBase:10, toBASE:_a[_i]}; var _r = this.convert("",10,_o.toBASE)
_s += "\r\n\oBASE.convert(n, '"+_o.fromBase+"', '"+_o.toBASE+"') ["+this.fromSymbols+"] ["+this.toSymbols+"]\r\n"
for(var _n=0;_n<(this.fromSymbols.length+2);_n++) { // iterate numbers
_r = this.convert(String(_n),_o.fromBase,_o.toBASE)
_s += _n+(String(_n)==this.convert(_r,_o.toBASE,_o.fromBase)?">":"?")+"["+_r+"] ";
}
}
return _s
}
var add = function(x, y, base) {
var _m = Math.max(x.length, y.length);
for(var _c = _n = 0,_r = []; _n < _m || _c; _c = Math.floor(_z / base)) {
var _z = _c + (_n < x.length ? x[_n] : 0) + (_n < y.length ? y[_n] : 0);
var _n = _r.push(_z % base);
}
return _r;
}
var mul = function(x, pow, base) {
for(var _r = x < 0 ? null : []; x > 0; x = x >> 1) {
if(x & 1) _r = add(_r, pow, base);
pow = add(pow, pow, base);
}
return _r;
}
}
Usage:
// quick test, convert from base45 to base32, using custom symbols for base85 and back to base45
var oBASE = new BASE();
var n = "THIS IS A NUMBER"; // Base 45 code = 'qr-alnum'
console.log(n); // Result: 'THIS IS A NUMBER'
var n = oBASE.convert(n,"qr-alnum",32); // Base 45 to Base 32 = 'base32hex'
console.log(n); // Result: '4ONI84LCTLJ1U08G1N'
var s85 = oBASE.rng("[0-9][a-z][A-Z].-:+=^!/*?&<>()[]{}#%$#"); // 32/Z85 custom symbols
var n = oBASE.convert(n,"base32hex",[s85]); // 'base2hex' to custom Base 85
console.log(n); // Result: 'fnaxrZP)?5d[DG'
var n = oBASE.convert(n,[s85],45); // Custom Base 85 to Base 45 = 'qr-alnum'
console.log(n); // Result: 'THIS IS A NUMBER'
function BASE(){this.convert=function(o,r,n){this.fromSymbols="object"==typeof r?r[0]:this.getsymbols(r),this.toSymbols="object"==typeof n?n[0]:this.getsymbols(n),r=this.fromSymbols.length,n=this.toSymbols.length;var i=[0];for(o=o.split("");o.length>0&&i[i.push(this.fromSymbols.indexOf(o.pop()))-1]>=0;);var h=i.shift()+i[i.length-1]>=0?i:null;if(null===h)return null;for(var e=0,l=(i=[],[1]);e<h.length;e++)i=t(i,s(h[e],l,n),n),l=s(r,l,n);e=i.length-1;for(var m="";e>=0;m+=this.toSymbols[i[e--]]);return 0==m.length?this.toSymbols[0]:m},this.symbols={32:function(){return this.base32hex},36:["[0-9][A-Z]"],45:function(){return this["qr-alnum"]},58:function(){return this.Bitcoin},64:["[A-Z][a-z][0-9]+/"],85:function(){return this["RFC 1924"]},91:['[A-Z][a-z][0-9]!#$%&()*+,./:;<=>?#[]^_`{|}~"'],94:["[!-~]"],geohash:["[0-9][b-h]jkmn[p-z]"],"RFC 4648":["[A-Z][2-7]"],base32hex:["[0-9][A-V]"],"qr-alnum":["[0-9][A-Z] $%*+-./:"],Bitcoin:["[1-9][A-H]JKLMN[P-Z][a-k][m-z]"],"RFC 1924":["[0-9][A-Z][a-z]!#$%&()*+-;<=>?#^_`{|}~"]},this.getsymbols=function(t){return void 0===this.symbols[t]&&(this.symbols[t]=t<95?this.rng(t<64?"[0-9][A-Z][a-z]+":"[A-Z][a-z][0-9][!-/][:-#][[-`][{-~]").substring(0,t):this.rng("[\0-ÿ]").substring(256-t,256)),"function"==typeof this.symbols[t]&&(this.symbols[t]=this.symbols[t]()),"object"==typeof this.symbols[t]&&(this.symbols[t]=this.rng(this.symbols[t][0])),this.symbols[t]},this.rng=function(t){var s=t.match(/\[.-.\]/);return null==s?t:(s=[s[0].charCodeAt(1),s[0].charCodeAt(3)],this.rng(t.replace(RegExp("\\[(\\x"+("0"+s[0].toString(16)).slice(-2)+"-\\x"+s[1].toString(16)+")\\]","g"),String.fromCharCode(..." ".repeat(s[1]-s[0]+1).split("").map((t,o)=>o+s[0])))))},this.selftest=function(){var t={};for(var s in this.symbols)t[s]=this.getsymbols(s).length;for(s=2;s<=95;s++)t[s]=this.getsymbols(s).length;t[256]=256;var o="";t=Object.keys(t).sort(function(s,o){return t[s]-t[o]});for(var r in t){s={fromBase:10,toBASE:t[r]};var n=this.convert("",10,s.toBASE);o+="\r\noBASE.convert(n, '"+s.fromBase+"', '"+s.toBASE+"') ["+this.fromSymbols+"] ["+this.toSymbols+"]\r\n";for(var i=0;i<this.fromSymbols.length+2;i++)n=this.convert(String(i),s.fromBase,s.toBASE),o+=i+(String(i)==this.convert(n,s.toBASE,s.fromBase)?">":"?")+"["+n+"] "}return o};var t=function(t,s,o){for(var r=Math.max(t.length,s.length),n=e=0,i=[];e<r||n;n=Math.floor(h/o))var h=n+(e<t.length?t[e]:0)+(e<s.length?s[e]:0),e=i.push(h%o);return i},s=function(s,o,r){for(var n=s<0?null:[];s>0;s>>=1)1&s&&(n=t(n,o,r)),o=t(o,o,r);return n}}
Self-Test:
// quick test, convert from base45 to base32, using custom symbols for base85 and back to base45
var oBASE = new BASE();
console.log(oBASE.selftest())
function BASE(){this.convert=function(o,r,n){this.fromSymbols="object"==typeof r?r[0]:this.getsymbols(r),this.toSymbols="object"==typeof n?n[0]:this.getsymbols(n),r=this.fromSymbols.length,n=this.toSymbols.length;var i=[0];for(o=o.split("");o.length>0&&i[i.push(this.fromSymbols.indexOf(o.pop()))-1]>=0;);var h=i.shift()+i[i.length-1]>=0?i:null;if(null===h)return null;for(var e=0,l=(i=[],[1]);e<h.length;e++)i=t(i,s(h[e],l,n),n),l=s(r,l,n);e=i.length-1;for(var m="";e>=0;m+=this.toSymbols[i[e--]]);return 0==m.length?this.toSymbols[0]:m},this.symbols={32:function(){return this.base32hex},36:["[0-9][A-Z]"],45:function(){return this["qr-alnum"]},58:function(){return this.Bitcoin},64:["[A-Z][a-z][0-9]+/"],85:function(){return this["RFC 1924"]},91:['[A-Z][a-z][0-9]!#$%&()*+,./:;<=>?#[]^_`{|}~"'],94:["[!-~]"],geohash:["[0-9][b-h]jkmn[p-z]"],"RFC 4648":["[A-Z][2-7]"],base32hex:["[0-9][A-V]"],"qr-alnum":["[0-9][A-Z] $%*+-./:"],Bitcoin:["[1-9][A-H]JKLMN[P-Z][a-k][m-z]"],"RFC 1924":["[0-9][A-Z][a-z]!#$%&()*+-;<=>?#^_`{|}~"]},this.getsymbols=function(t){return void 0===this.symbols[t]&&(this.symbols[t]=t<95?this.rng(t<64?"[0-9][A-Z][a-z]+":"[A-Z][a-z][0-9][!-/][:-#][[-`][{-~]").substring(0,t):this.rng("[\0-ÿ]").substring(256-t,256)),"function"==typeof this.symbols[t]&&(this.symbols[t]=this.symbols[t]()),"object"==typeof this.symbols[t]&&(this.symbols[t]=this.rng(this.symbols[t][0])),this.symbols[t]},this.rng=function(t){var s=t.match(/\[.-.\]/);return null==s?t:(s=[s[0].charCodeAt(1),s[0].charCodeAt(3)],this.rng(t.replace(RegExp("\\[(\\x"+("0"+s[0].toString(16)).slice(-2)+"-\\x"+s[1].toString(16)+")\\]","g"),String.fromCharCode(..." ".repeat(s[1]-s[0]+1).split("").map((t,o)=>o+s[0])))))},this.selftest=function(){var t={};for(var s in this.symbols)t[s]=this.getsymbols(s).length;for(s=2;s<=95;s++)t[s]=this.getsymbols(s).length;t[256]=256;var o="";t=Object.keys(t).sort(function(s,o){return t[s]-t[o]});for(var r in t){s={fromBase:10,toBASE:t[r]};var n=this.convert("",10,s.toBASE);o+="\r\noBASE.convert(n, '"+s.fromBase+"', '"+s.toBASE+"') ["+this.fromSymbols+"] ["+this.toSymbols+"]\r\n";for(var i=0;i<this.fromSymbols.length+2;i++)n=this.convert(String(i),s.fromBase,s.toBASE),o+=i+(String(i)==this.convert(n,s.toBASE,s.fromBase)?">":"?")+"["+n+"] "}return o};var t=function(t,s,o){for(var r=Math.max(t.length,s.length),n=e=0,i=[];e<r||n;n=Math.floor(h/o))var h=n+(e<t.length?t[e]:0)+(e<s.length?s[e]:0),e=i.push(h%o);return i},s=function(s,o,r){for(var n=s<0?null:[];s>0;s>>=1)1&s&&(n=t(n,o,r)),o=t(o,o,r);return n}}

to convert numbers into different bases in JavaScript or typescript using the following ways.
function binary(number) {
console.log((number >>> 0).toString(2)); //base 2 for binary
}
binary(-7);
function octal(number) {
console.log(number.toString(8)); //base 8 for octal
}
octal(15);
function hex(number) {
console.log(number.toString(16)); //base 16 for hex
}
hex(15);
here in the binary function, you can use number.toString(2) function, but the problem occurs when representing negative numbers. so that you can use the unsigned right shift bitwise operator (>>>) to fix this issue.

You can also convert number in hexadecimal to decimal as follows:
var a="8F";
var b=a.split("");
var result=0;var hex_multiplier=1;
for(var i=0;i<b.length;i++){
result +=parseInt(b[i],16)*hex_multiplier;
hex_multiplier *=16;
}
console.log(result);
where you can change a with any hexadecimal number and get the result in decimal form.

You can use JavaScript's built-in integer literals for some scenarios:
function binaryToDecimal(binaryString) {
return Number('0b' + binaryString.replace('-', '')) * signOf(binaryString);;
}
function octalToDecimal(octalString) {
return Number('0o' + octalString.replace('-', '')) * signOf(octalString);
}
function hexToDecimal(hexString) {
return Number('0x' + hexString.replace('-', '')) * signOf(hexString);
}
function signOf(n) {
return n.trim()[0] == '-' ? -1 : 1;
}
console.log(binaryToDecimal('-0101'),
octalToDecimal('-052171'),
hexToDecimal('deadbeef'));

This code converts a number to an arbitrary-large base, outputting an array of digits.
function convertBase(num, base) {
var result = [];
while(num >= 1) {
result.unshift(num % base);
num = Math.floor(num / base);
}
return result;
}
console.log(convertBase(100, 12)); // [8, 4]
console.log(convertBase(5114, 64)); // [1, 15, 58]

TypeScript Implementation that supports decimal and int:
class Digit {
private readonly digitMap: Map<string, number>;
constructor(readonly digits: string, readonly maxDigits = 10) {
this.digitMap = new Map<string, number>();
for (let i = 0; i < digits.length; i += 1) {
if (digits[i] === ".") {
throw new TypeError(`Digits should not contain the "." mark.`);
}
if (this.digitMap.has(digits[i])) {
throw new TypeError(`Duplicated digit character.`);
}
this.digitMap.set(digits[i], i);
}
}
to = (x: number) => {
const int = Math.floor(x);
const dec = x - int;
let intRemains = int;
let intResult = "";
while (true) {
intResult = this.digits[intRemains % this.digits.length] + intResult;
intRemains = Math.floor(intRemains / this.digits.length);
if (intRemains === 0) break;
}
if (dec > 0) {
let decRemains = dec;
let decResult = "";
for (let i = 0; i < this.maxDigits; i += 1) {
const stepCache = decRemains * this.digits.length;
const decInt = Math.floor(stepCache);
decResult = decResult + this.digits[decInt];
decRemains = stepCache - decInt;
if (decRemains === 0) break;
}
if (decResult === "0") return intResult;
return intResult + "." + decResult;
}
return intResult;
};
from = (x: string) => {
const splitted = x.split(".");
if (splitted.length > 2) return Number.NaN;
const [int, dec] = splitted;
let result = 0;
for (let i = 0; i < int.length; i += 1) {
const digit = int[int.length - i - 1];
const oct = this.digitMap.get(digit);
if (oct === undefined) return Number.NaN;
result += oct * Math.pow(this.digits.length, i);
}
if (dec) {
for (let i = 0; i < dec.length; i += 1) {
const digit = dec[i];
const oct = this.digitMap.get(digit);
if (oct === undefined) return Number.NaN;
result += oct * Math.pow(this.digits.length, 0 - i - 1);
}
}
return result;
};
}
Here're the example about how to use it:
Creating the converter:
const d = new Digit("0123456789");
Convert a number to string:
d.to(1.2345);
Convert a string to number:
d.from("1.2345");
You can also set the second parameter to control the maximum number of decimal places, for example:
const d0 = new Digit("abcdABCD", 10);
d0.to(0.96);
// a.DBdAbcbDcD
const d1 = new Digit("abcdABCD", 20);
d1.to(0.96);
// a.DBdAbcbDcDacAdCBC

Monkey patched Number prototype for all radixes up to and including 64
( inspired by and derived from #nirvana's answer )
NumberToStringPatched.__patched = NumberToStringPatched;
if (Number.prototype.toString.__patched!==NumberToStringPatched) {
NumberToStringPatched.__unpatched = Number.prototype.toString;
Number.prototype.toString = NumberToStringPatched;
}
NumberParseIntPatched.__patched = NumberParseIntPatched;
if (Number.parseInt.__patched!==NumberParseIntPatched) {
NumberParseIntPatched.__unpatched = Number.parseInt;
Number.parseInt = NumberParseIntPatched;
}
function NumberToStringPatched(base=10) {
if (base<33||base>64) return NumberToStringPatched.__unpatched.call(this,base);
return convertToBase(this,base);
}
function NumberParseIntPatched(str,base=10) {
if (base<33||base>64) return NumberParseIntPatched.__unpatched.call(this,str,base);
return convertFromBase(str,base);
}
const numberCharSet=(()=>{
let chars = "0123456789";
for (let c='a';c<='z';c=String.fromCharCode(c.charCodeAt(0)+1)) {
chars+=c;
}
for (let c='A';c<='Z';c=String.fromCharCode(c.charCodeAt(0)+1)) {
chars+=c;
}
return (chars+'_$').split('');
})();
function convertToBase(num, base) {
var result = '';
while(num >= 1) {
result = numberCharSet[num % base]+result;
num = Math.floor(num / base);
}
return result;
}
function convertFromBase(str, base) {
let charset = numberCharSet.slice(0,base);
if (typeof str==='number') str=str.toString(base);
if (base <= 26) str=str.toLowerCase();
let digits = str.split('').map(function(c){
let x = charset.indexOf(c);
if (x<0) throw new Error("invalid digit for given radix "+base+': '+c);
return x;
});
let result = digits.shift();
while (digits.length) {
result=(result*base)+digits.shift();
}
return result;
}
for (let i = 0;i<64;i++) {
console.log(i.toString(64));
}
console.log(9000);
console.log(9 * 10 * 10 * 10);
console.log(0xf000);
console.log(15 * 16 * 16 * 16);
console.log(Number.parseInt('$000',64));
console.log(63 * 64 * 64 * 64);

I am writing this answer simply because I was kinda stupid to understand the difference between parseInt and .toString().
Here is the difference
parseInt
parseInt(str, base) will convert the str to an Integer and base here is used to tell parseInt which base the str is in. I thought the base here was what the str would get converted to. But thats not true. parseInt is just used to convert a number in any base to an integer in base10.
.toString()
number.toString(base) is used to convert the number into a number with base base and it assumes that number is in Integer form. So if the number is in hex format and you want to convert it to binary you need to convert the number first to Integer using parseInt and then you can use this function to change the base.
Hope this helps if you are as stupid as me. :)

Your Own recursion method for creating base2 conversions.
As mentioned by the users above
let n = 13; console.log(n.toString(2)); will result in 13 conversion from base 10 to base 2.
But in case if you want to program the same. I have written a recursive method to do the same. which just simply divide by 2 and then count remainders.
// #author Tarandeep Singh :: Created recursive converter from base 10 to base 2
// #date : 2017-04-11
// Convert Base 10 to Base 2, We should reverse the output
// For Example base10to2(10) = "0101" just do res = base10to2(10).split('').reverse().join();
function base10to2(val, res = '') {
if (val >= 2) {
res += '' + val % 2;
return base10to2(val = Math.floor(val / 2), res);
} else {
res += '' + 1
return res;
}
}
let n = 13;
var result = base10to2(n).split('').reverse().join();
document.write(`Converting ${n} into Base2 is ${result}`);

which are alternative of tofixed() in javascript [duplicate]

Suppose I have a value of 15.7784514, I want to display it 15.77 with no rounding.
var num = parseFloat(15.7784514);
document.write(num.toFixed(1)+"<br />");
document.write(num.toFixed(2)+"<br />");
document.write(num.toFixed(3)+"<br />");
document.write(num.toFixed(10));
Results in -
15.8
15.78
15.778
15.7784514000
How do I display 15.77?

Convert the number into a string, match the number up to the second decimal place:
function calc(theform) {
var num = theform.original.value, rounded = theform.rounded
var with2Decimals = num.toString().match(/^-?\d+(?:\.\d{0,2})?/)[0]
rounded.value = with2Decimals
}
<form onsubmit="return calc(this)">
Original number: <input name="original" type="text" onkeyup="calc(form)" onchange="calc(form)" />
<br />"Rounded" number: <input name="rounded" type="text" placeholder="readonly" readonly>
</form>
The toFixed method fails in some cases unlike toString, so be very careful with it.

Update 5 Nov 2016
New answer, always accurate
function toFixed(num, fixed) {
var re = new RegExp('^-?\\d+(?:\.\\d{0,' + (fixed || -1) + '})?');
return num.toString().match(re)[0];
}
As floating point math in javascript will always have edge cases, the previous solution will be accurate most of the time which is not good enough.
There are some solutions to this like num.toPrecision, BigDecimal.js, and accounting.js.
Yet, I believe that merely parsing the string will be the simplest and always accurate.
Basing the update on the well written regex from the accepted answer by #Gumbo, this new toFixed function will always work as expected.
Old answer, not always accurate.
Roll your own toFixed function:
function toFixed(num, fixed) {
fixed = fixed || 0;
fixed = Math.pow(10, fixed);
return Math.floor(num * fixed) / fixed;
}

Another single-line solution :
number = Math.trunc(number*100)/100
I used 100 because you want to truncate to the second digit, but a more flexible solution would be :
number = Math.trunc(number*Math.pow(10, digits))/Math.pow(10, digits)
where digits is the amount of decimal digits to keep.
See Math.trunc specs for details and browser compatibility.

I opted to write this instead to manually remove the remainder with strings so I don't have to deal with the math issues that come with numbers:
num = num.toString(); //If it's not already a String
num = num.slice(0, (num.indexOf("."))+3); //With 3 exposing the hundredths place
Number(num); //If you need it back as a Number
This will give you "15.77" with num = 15.7784514;

Update (Jan 2021)
Depending on its range, a number in javascript may be shown in scientific notation. For example, if you type 0.0000001 in the console, you may see it as 1e-7, whereas 0.000001 appears unchanged (0.000001).
If your application works on a range of numbers for which scientific notation is not involved, you can just ignore this update and use the original answer below.
This update is about adding a function that checks if the number is in scientific format and, if so, converts it into decimal format. Here I'm proposing this one, but you can use any other function that achieves the same goal, according to your application's needs:
function toFixed(x) {
if (Math.abs(x) < 1.0) {
let e = parseInt(x.toString().split('e-')[1]);
if (e) {
x *= Math.pow(10,e-1);
x = '0.' + (new Array(e)).join('0') + x.toString().substring(2);
}
} else {
let e = parseInt(x.toString().split('+')[1]);
if (e > 20) {
e -= 20;
x /= Math.pow(10,e);
x += (new Array(e+1)).join('0');
}
}
return x;
}
Now just apply that function to the parameter (that's the only change with respect to the original answer):
function toFixedTrunc(x, n) {
x = toFixed(x)
// From here on the code is the same than the original answer
const v = (typeof x === 'string' ? x : x.toString()).split('.');
if (n <= 0) return v[0];
let f = v[1] || '';
if (f.length > n) return `${v[0]}.${f.substr(0,n)}`;
while (f.length < n) f += '0';
return `${v[0]}.${f}`
}
This updated version addresses also a case mentioned in a comment:
toFixedTrunc(0.000000199, 2) => "0.00"
Again, choose what fits your application needs at best.
Original answer (October 2017)
General solution to truncate (no rounding) a number to the n-th decimal digit and convert it to a string with exactly n decimal digits, for any n≥0.
function toFixedTrunc(x, n) {
const v = (typeof x === 'string' ? x : x.toString()).split('.');
if (n <= 0) return v[0];
let f = v[1] || '';
if (f.length > n) return `${v[0]}.${f.substr(0,n)}`;
while (f.length < n) f += '0';
return `${v[0]}.${f}`
}
where x can be either a number (which gets converted into a string) or a string.
Here are some tests for n=2 (including the one requested by OP):
0 => 0.00
0.01 => 0.01
0.5839 => 0.58
0.999 => 0.99
1.01 => 1.01
2 => 2.00
2.551 => 2.55
2.99999 => 2.99
4.27 => 4.27
15.7784514 => 15.77
123.5999 => 123.59
And for some other values of n:
15.001097 => 15.0010 (n=4)
0.000003298 => 0.0000032 (n=7)
0.000003298257899 => 0.000003298257 (n=12)

parseInt is faster then Math.floor
function floorFigure(figure, decimals){
if (!decimals) decimals = 2;
var d = Math.pow(10,decimals);
return (parseInt(figure*d)/d).toFixed(decimals);
};
floorFigure(123.5999) => "123.59"
floorFigure(123.5999, 3) => "123.599"

num = 19.66752
f = num.toFixed(3).slice(0,-1)
alert(f)
This will return 19.66

Simple do this
number = parseInt(number * 100)/100;

Just truncate the digits:
function truncDigits(inputNumber, digits) {
const fact = 10 ** digits;
return Math.floor(inputNumber * fact) / fact;
}

This is not a safe alternative, as many others commented examples with numbers that turn into exponential notation, that scenery is not covered by this function
// typescript
// function formatLimitDecimals(value: number, decimals: number): number {
function formatLimitDecimals(value, decimals) {
const stringValue = value.toString();
if(stringValue.includes('e')) {
// TODO: remove exponential notation
throw 'invald number';
} else {
const [integerPart, decimalPart] = stringValue.split('.');
if(decimalPart) {
return +[integerPart, decimalPart.slice(0, decimals)].join('.')
} else {
return integerPart;
}
}
}
console.log(formatLimitDecimals(4.156, 2)); // 4.15
console.log(formatLimitDecimals(4.156, 8)); // 4.156
console.log(formatLimitDecimals(4.156, 0)); // 4
console.log(formatLimitDecimals(0, 4)); // 0
// not covered
console.log(formatLimitDecimals(0.000000199, 2)); // 0.00

These solutions do work, but to me seem unnecessarily complicated. I personally like to use the modulus operator to obtain the remainder of a division operation, and remove that. Assuming that num = 15.7784514:
num-=num%.01;
This is equivalent to saying num = num - (num % .01).

I fixed using following simple way-
var num = 15.7784514;
Math.floor(num*100)/100;
Results will be 15.77

My version for positive numbers:
function toFixed_norounding(n,p)
{
var result = n.toFixed(p);
return result <= n ? result: (result - Math.pow(0.1,p)).toFixed(p);
}
Fast, pretty, obvious. (version for positive numbers)

The answers here didn't help me, it kept rounding up or giving me the wrong decimal.
my solution converts your decimal to a string, extracts the characters and then returns the whole thing as a number.
function Dec2(num) {
num = String(num);
if(num.indexOf('.') !== -1) {
var numarr = num.split(".");
if (numarr.length == 1) {
return Number(num);
}
else {
return Number(numarr[0]+"."+numarr[1].charAt(0)+numarr[1].charAt(1));
}
}
else {
return Number(num);
}
}
Dec2(99); // 99
Dec2(99.9999999); // 99.99
Dec2(99.35154); // 99.35
Dec2(99.8); // 99.8
Dec2(10265.985475); // 10265.98

The following code works very good for me:
num.toString().match(/.\*\\..{0,2}|.\*/)[0];

This worked well for me. I hope it will fix your issues too.
function toFixedNumber(number) {
const spitedValues = String(number.toLocaleString()).split('.');
let decimalValue = spitedValues.length > 1 ? spitedValues[1] : '';
decimalValue = decimalValue.concat('00').substr(0,2);
return '$'+spitedValues[0] + '.' + decimalValue;
}
// 5.56789 ----> $5.56
// 0.342 ----> $0.34
// -10.3484534 ----> $-10.34
// 600 ----> $600.00
function convertNumber(){
var result = toFixedNumber(document.getElementById("valueText").value);
document.getElementById("resultText").value = result;
}
function toFixedNumber(number) {
const spitedValues = String(number.toLocaleString()).split('.');
let decimalValue = spitedValues.length > 1 ? spitedValues[1] : '';
decimalValue = decimalValue.concat('00').substr(0,2);
return '$'+spitedValues[0] + '.' + decimalValue;
}
<div>
<input type="text" id="valueText" placeholder="Input value here..">
<br>
<button onclick="convertNumber()" >Convert</button>
<br><hr>
<input type="text" id="resultText" placeholder="result" readonly="true">
</div>

An Easy way to do it is the next but is necessary ensure that the amount parameter is given as a string.
function truncate(amountAsString, decimals = 2){
var dotIndex = amountAsString.indexOf('.');
var toTruncate = dotIndex !== -1 && ( amountAsString.length > dotIndex + decimals + 1);
var approach = Math.pow(10, decimals);
var amountToTruncate = toTruncate ? amountAsString.slice(0, dotIndex + decimals +1) : amountAsString;
return toTruncate
? Math.floor(parseFloat(amountToTruncate) * approach ) / approach
: parseFloat(amountAsString);
}
console.log(truncate("7.99999")); //OUTPUT ==> 7.99
console.log(truncate("7.99999", 3)); //OUTPUT ==> 7.999
console.log(truncate("12.799999999999999")); //OUTPUT ==> 7.99

Here you are. An answer that shows yet another way to solve the problem:
// For the sake of simplicity, here is a complete function:
function truncate(numToBeTruncated, numOfDecimals) {
var theNumber = numToBeTruncated.toString();
var pointIndex = theNumber.indexOf('.');
return +(theNumber.slice(0, pointIndex > -1 ? ++numOfDecimals + pointIndex : undefined));
}
Note the use of + before the final expression. That is to convert our truncated, sliced string back to number type.
Hope it helps!

truncate without zeroes
function toTrunc(value,n){
return Math.floor(value*Math.pow(10,n))/(Math.pow(10,n));
}
or
function toTrunc(value,n){
x=(value.toString()+".0").split(".");
return parseFloat(x[0]+"."+x[1].substr(0,n));
}
test:
toTrunc(17.4532,2) //17.45
toTrunc(177.4532,1) //177.4
toTrunc(1.4532,1) //1.4
toTrunc(.4,2) //0.4
truncate with zeroes
function toTruncFixed(value,n){
return toTrunc(value,n).toFixed(n);
}
test:
toTrunc(17.4532,2) //17.45
toTrunc(177.4532,1) //177.4
toTrunc(1.4532,1) //1.4
toTrunc(.4,2) //0.40

If you exactly wanted to truncate to 2 digits of precision, you can go with a simple logic:
function myFunction(number) {
var roundedNumber = number.toFixed(2);
if (roundedNumber > number)
{
roundedNumber = roundedNumber - 0.01;
}
return roundedNumber;
}

I used (num-0.05).toFixed(1) to get the second decimal floored.

It's more reliable to get two floating points without rounding.
Reference Answer
var number = 10.5859;
var fixed2FloatPoints = parseInt(number * 100) / 100;
console.log(fixed2FloatPoints);
Thank You !

My solution in typescript (can easily be ported to JS):
/**
* Returns the price with correct precision as a string
*
* #param price The price in decimal to be formatted.
* #param decimalPlaces The number of decimal places to use
* #return string The price in Decimal formatting.
*/
type toDecimal = (price: number, decimalPlaces?: number) => string;
const toDecimalOdds: toDecimal = (
price: number,
decimalPlaces: number = 2,
): string => {
const priceString: string = price.toString();
const pointIndex: number = priceString.indexOf('.');
// Return the integer part if decimalPlaces is 0
if (decimalPlaces === 0) {
return priceString.substr(0, pointIndex);
}
// Return value with 0s appended after decimal if the price is an integer
if (pointIndex === -1) {
const padZeroString: string = '0'.repeat(decimalPlaces);
return `${priceString}.${padZeroString}`;
}
// If numbers after decimal are less than decimalPlaces, append with 0s
const padZeroLen: number = priceString.length - pointIndex - 1;
if (padZeroLen > 0 && padZeroLen < decimalPlaces) {
const padZeroString: string = '0'.repeat(padZeroLen);
return `${priceString}${padZeroString}`;
}
return priceString.substr(0, pointIndex + decimalPlaces + 1);
};
Test cases:
expect(filters.toDecimalOdds(3.14159)).toBe('3.14');
expect(filters.toDecimalOdds(3.14159, 2)).toBe('3.14');
expect(filters.toDecimalOdds(3.14159, 0)).toBe('3');
expect(filters.toDecimalOdds(3.14159, 10)).toBe('3.1415900000');
expect(filters.toDecimalOdds(8.2)).toBe('8.20');
Any improvements?

Another solution, that truncates and round:
function round (number, decimals, truncate) {
if (truncate) {
number = number.toFixed(decimals + 1);
return parseFloat(number.slice(0, -1));
}
var n = Math.pow(10.0, decimals);
return Math.round(number * n) / n;
};

function limitDecimalsWithoutRounding(val, decimals){
let parts = val.toString().split(".");
return parseFloat(parts[0] + "." + parts[1].substring(0, decimals));
}
var num = parseFloat(15.7784514);
var new_num = limitDecimalsWithoutRounding(num, 2);

Roll your own toFixed function: for positive values Math.floor works fine.
function toFixed(num, fixed) {
fixed = fixed || 0;
fixed = Math.pow(10, fixed);
return Math.floor(num * fixed) / fixed;
}
For negative values Math.floor is round of the values. So you can use Math.ceil instead.
Example,
Math.ceil(-15.778665 * 10000) / 10000 = -15.7786
Math.floor(-15.778665 * 10000) / 10000 = -15.7787 // wrong.

Gumbo's second solution, with the regular expression, does work but is slow because of the regular expression. Gumbo's first solution fails in certain situations due to imprecision in floating points numbers. See the JSFiddle for a demonstration and a benchmark. The second solution takes about 1636 nanoseconds per call on my current system, Intel Core i5-2500 CPU at 3.30 GHz.
The solution I've written involves adding a small compensation to take care of floating point imprecision. It is basically instantaneous, i.e. on the order of nanoseconds. I clocked 2 nanoseconds per call but the JavaScript timers are not very precise or granular. Here is the JS Fiddle and the code.
function toFixedWithoutRounding (value, precision)
{
var factorError = Math.pow(10, 14);
var factorTruncate = Math.pow(10, 14 - precision);
var factorDecimal = Math.pow(10, precision);
return Math.floor(Math.floor(value * factorError + 1) / factorTruncate) / factorDecimal;
}
var values = [1.1299999999, 1.13, 1.139999999, 1.14, 1.14000000001, 1.13 * 100];
for (var i = 0; i < values.length; i++)
{
var value = values[i];
console.log(value + " --> " + toFixedWithoutRounding(value, 2));
}
for (var i = 0; i < values.length; i++)
{
var value = values[i];
console.log(value + " --> " + toFixedWithoutRounding(value, 4));
}
console.log("type of result is " + typeof toFixedWithoutRounding(1.13 * 100 / 100, 2));
// Benchmark
var value = 1.13 * 100;
var startTime = new Date();
var numRun = 1000000;
var nanosecondsPerMilliseconds = 1000000;
for (var run = 0; run < numRun; run++)
toFixedWithoutRounding(value, 2);
var endTime = new Date();
var timeDiffNs = nanosecondsPerMilliseconds * (endTime - startTime);
var timePerCallNs = timeDiffNs / numRun;
console.log("Time per call (nanoseconds): " + timePerCallNs);

Building on David D's answer:
function NumberFormat(num,n) {
var num = (arguments[0] != null) ? arguments[0] : 0;
var n = (arguments[1] != null) ? arguments[1] : 2;
if(num > 0){
num = String(num);
if(num.indexOf('.') !== -1) {
var numarr = num.split(".");
if (numarr.length > 1) {
if(n > 0){
var temp = numarr[0] + ".";
for(var i = 0; i < n; i++){
if(i < numarr[1].length){
temp += numarr[1].charAt(i);
}
}
num = Number(temp);
}
}
}
}
return Number(num);
}
console.log('NumberFormat(123.85,2)',NumberFormat(123.85,2));
console.log('NumberFormat(123.851,2)',NumberFormat(123.851,2));
console.log('NumberFormat(0.85,2)',NumberFormat(0.85,2));
console.log('NumberFormat(0.851,2)',NumberFormat(0.851,2));
console.log('NumberFormat(0.85156,2)',NumberFormat(0.85156,2));
console.log('NumberFormat(0.85156,4)',NumberFormat(0.85156,4));
console.log('NumberFormat(0.85156,8)',NumberFormat(0.85156,8));
console.log('NumberFormat(".85156",2)',NumberFormat(".85156",2));
console.log('NumberFormat("0.85156",2)',NumberFormat("0.85156",2));
console.log('NumberFormat("1005.85156",2)',NumberFormat("1005.85156",2));
console.log('NumberFormat("0",2)',NumberFormat("0",2));
console.log('NumberFormat("",2)',NumberFormat("",2));
console.log('NumberFormat(85156,8)',NumberFormat(85156,8));
console.log('NumberFormat("85156",2)',NumberFormat("85156",2));
console.log('NumberFormat("85156.",2)',NumberFormat("85156.",2));
// NumberFormat(123.85,2) 123.85
// NumberFormat(123.851,2) 123.85
// NumberFormat(0.85,2) 0.85
// NumberFormat(0.851,2) 0.85
// NumberFormat(0.85156,2) 0.85
// NumberFormat(0.85156,4) 0.8515
// NumberFormat(0.85156,8) 0.85156
// NumberFormat(".85156",2) 0.85
// NumberFormat("0.85156",2) 0.85
// NumberFormat("1005.85156",2) 1005.85
// NumberFormat("0",2) 0
// NumberFormat("",2) 0
// NumberFormat(85156,8) 85156
// NumberFormat("85156",2) 85156
// NumberFormat("85156.",2) 85156

Already there are some suitable answer with regular expression and arithmetic calculation, you can also try this
function myFunction() {
var str = 12.234556;
str = str.toString().split('.');
var res = str[1].slice(0, 2);
document.getElementById("demo").innerHTML = str[0]+'.'+res;
}
// output: 12.23

Here is what is did it with string
export function withoutRange(number) {
const str = String(number);
const dotPosition = str.indexOf('.');
if (dotPosition > 0) {
const length = str.substring().length;
const end = length > 3 ? 3 : length;
return str.substring(0, dotPosition + end);
}
return str;
}

convert whole number to a decimal javascript [duplicate]

I have this line of code which rounds my numbers to two decimal places. But I get numbers like this: 10.8, 2.4, etc. These are not my idea of two decimal places so how I can improve the following?
Math.round(price*Math.pow(10,2))/Math.pow(10,2);
I want numbers like 10.80, 2.40, etc. Use of jQuery is fine with me.

To format a number using fixed-point notation, you can simply use the toFixed method:
(10.8).toFixed(2); // "10.80"
var num = 2.4;
alert(num.toFixed(2)); // "2.40"
Note that toFixed() returns a string.
IMPORTANT: Note that toFixed does not round 90% of the time, it will return the rounded value, but for many cases, it doesn't work.
For instance:
2.005.toFixed(2) === "2.00"
UPDATE:
Nowadays, you can use the Intl.NumberFormat constructor. It's part of the ECMAScript Internationalization API Specification (ECMA402). It has pretty good browser support, including even IE11, and it is fully supported in Node.js.
const formatter = new Intl.NumberFormat('en-US', {
minimumFractionDigits: 2,
maximumFractionDigits: 2,
});
console.log(formatter.format(2.005)); // "2.01"
console.log(formatter.format(1.345)); // "1.35"
You can alternatively use the toLocaleString method, which internally will use the Intl API:
const format = (num, decimals) => num.toLocaleString('en-US', {
minimumFractionDigits: 2,
maximumFractionDigits: 2,
});
console.log(format(2.005)); // "2.01"
console.log(format(1.345)); // "1.35"
This API also provides you a wide variety of options to format, like thousand separators, currency symbols, etc.

This is an old topic but still top-ranked Google results and the solutions offered share the same floating point decimals issue. Here is the (very generic) function I use, thanks to MDN:
function round(value, exp) {
if (typeof exp === 'undefined' || +exp === 0)
return Math.round(value);
value = +value;
exp = +exp;
if (isNaN(value) || !(typeof exp === 'number' && exp % 1 === 0))
return NaN;
// Shift
value = value.toString().split('e');
value = Math.round(+(value[0] + 'e' + (value[1] ? (+value[1] + exp) : exp)));
// Shift back
value = value.toString().split('e');
return +(value[0] + 'e' + (value[1] ? (+value[1] - exp) : -exp));
}
As we can see, we don't get these issues:
round(1.275, 2); // Returns 1.28
round(1.27499, 2); // Returns 1.27
This genericity also provides some cool stuff:
round(1234.5678, -2); // Returns 1200
round(1.2345678e+2, 2); // Returns 123.46
round("123.45"); // Returns 123
Now, to answer the OP's question, one has to type:
round(10.8034, 2).toFixed(2); // Returns "10.80"
round(10.8, 2).toFixed(2); // Returns "10.80"
Or, for a more concise, less generic function:
function round2Fixed(value) {
value = +value;
if (isNaN(value))
return NaN;
// Shift
value = value.toString().split('e');
value = Math.round(+(value[0] + 'e' + (value[1] ? (+value[1] + 2) : 2)));
// Shift back
value = value.toString().split('e');
return (+(value[0] + 'e' + (value[1] ? (+value[1] - 2) : -2))).toFixed(2);
}
You can call it with:
round2Fixed(10.8034); // Returns "10.80"
round2Fixed(10.8); // Returns "10.80"
Various examples and tests (thanks to #t-j-crowder!):
function round(value, exp) {
if (typeof exp === 'undefined' || +exp === 0)
return Math.round(value);
value = +value;
exp = +exp;
if (isNaN(value) || !(typeof exp === 'number' && exp % 1 === 0))
return NaN;
// Shift
value = value.toString().split('e');
value = Math.round(+(value[0] + 'e' + (value[1] ? (+value[1] + exp) : exp)));
// Shift back
value = value.toString().split('e');
return +(value[0] + 'e' + (value[1] ? (+value[1] - exp) : -exp));
}
function naive(value, exp) {
if (!exp) {
return Math.round(value);
}
var pow = Math.pow(10, exp);
return Math.round(value * pow) / pow;
}
function test(val, places) {
subtest(val, places);
val = typeof val === "string" ? "-" + val : -val;
subtest(val, places);
}
function subtest(val, places) {
var placesOrZero = places || 0;
var naiveResult = naive(val, places);
var roundResult = round(val, places);
if (placesOrZero >= 0) {
naiveResult = naiveResult.toFixed(placesOrZero);
roundResult = roundResult.toFixed(placesOrZero);
} else {
naiveResult = naiveResult.toString();
roundResult = roundResult.toString();
}
$("<tr>")
.append($("<td>").text(JSON.stringify(val)))
.append($("<td>").text(placesOrZero))
.append($("<td>").text(naiveResult))
.append($("<td>").text(roundResult))
.appendTo("#results");
}
test(0.565, 2);
test(0.575, 2);
test(0.585, 2);
test(1.275, 2);
test(1.27499, 2);
test(1234.5678, -2);
test(1.2345678e+2, 2);
test("123.45");
test(10.8034, 2);
test(10.8, 2);
test(1.005, 2);
test(1.0005, 2);
table {
border-collapse: collapse;
}
table, td, th {
border: 1px solid #ddd;
}
td, th {
padding: 4px;
}
th {
font-weight: normal;
font-family: sans-serif;
}
td {
font-family: monospace;
}
<table>
<thead>
<tr>
<th>Input</th>
<th>Places</th>
<th>Naive</th>
<th>Thorough</th>
</tr>
</thead>
<tbody id="results">
</tbody>
</table>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

I usually add this to my personal library, and after some suggestions and using the #TIMINeutron solution too, and making it adaptable for decimal length then, this one fits best:
function precise_round(num, decimals) {
var t = Math.pow(10, decimals);
return (Math.round((num * t) + (decimals>0?1:0)*(Math.sign(num) * (10 / Math.pow(100, decimals)))) / t).toFixed(decimals);
}
will work for the exceptions reported.

FAST AND EASY
parseFloat(number.toFixed(2))
Example
let number = 2.55435930
let roundedString = number.toFixed(2) // "2.55"
let twoDecimalsNumber = parseFloat(roundedString) // 2.55
let directly = parseFloat(number.toFixed(2)) // 2.55

One way to be 100% sure that you get a number with 2 decimals:
(Math.round(num*100)/100).toFixed(2)
If this causes rounding errors, you can use the following as James has explained in his comment:
(Math.round((num * 1000)/10)/100).toFixed(2)

I don't know why can't I add a comment to a previous answer (maybe I'm hopelessly blind, I don't know), but I came up with a solution using #Miguel's answer:
function precise_round(num,decimals) {
return Math.round(num*Math.pow(10, decimals)) / Math.pow(10, decimals);
}
And its two comments (from #bighostkim and #Imre):
Problem with precise_round(1.275,2) not returning 1.28
Problem with precise_round(6,2) not returning 6.00 (as he wanted).
My final solution is as follows:
function precise_round(num,decimals) {
var sign = num >= 0 ? 1 : -1;
return (Math.round((num*Math.pow(10,decimals)) + (sign*0.001)) / Math.pow(10,decimals)).toFixed(decimals);
}
As you can see I had to add a little bit of "correction" (it's not what it is, but since Math.round is lossy - you can check it on jsfiddle.net - this is the only way I knew how to "fix" it). It adds 0.001 to the already padded number, so it is adding a 1 three 0s to the right of the decimal value. So it should be safe to use.
After that I added .toFixed(decimal) to always output the number in the correct format (with the right amount of decimals).
So that's pretty much it. Use it well ;)
EDIT: added functionality to the "correction" of negative numbers.

toFixed(n) provides n length after the decimal point; toPrecision(x)
provides x total length.
Use this method below
// Example: toPrecision(4) when the number has 7 digits (3 before, 4 after)
// It will round to the tenths place
num = 500.2349;
result = num.toPrecision(4); // result will equal 500.2
AND if you want the number to be fixed use
result = num.toFixed(2);

I didn't find an accurate solution for this problem, so I created my own:
function inprecise_round(value, decPlaces) {
return Math.round(value*Math.pow(10,decPlaces))/Math.pow(10,decPlaces);
}
function precise_round(value, decPlaces){
var val = value * Math.pow(10, decPlaces);
var fraction = (Math.round((val-parseInt(val))*10)/10);
//this line is for consistency with .NET Decimal.Round behavior
// -342.055 => -342.06
if(fraction == -0.5) fraction = -0.6;
val = Math.round(parseInt(val) + fraction) / Math.pow(10, decPlaces);
return val;
}
Examples:
function inprecise_round(value, decPlaces) {
return Math.round(value * Math.pow(10, decPlaces)) / Math.pow(10, decPlaces);
}
function precise_round(value, decPlaces) {
var val = value * Math.pow(10, decPlaces);
var fraction = (Math.round((val - parseInt(val)) * 10) / 10);
//this line is for consistency with .NET Decimal.Round behavior
// -342.055 => -342.06
if (fraction == -0.5) fraction = -0.6;
val = Math.round(parseInt(val) + fraction) / Math.pow(10, decPlaces);
return val;
}
// This may produce different results depending on the browser environment
console.log("342.055.toFixed(2) :", 342.055.toFixed(2)); // 342.06 on Chrome & IE10
console.log("inprecise_round(342.055, 2):", inprecise_round(342.055, 2)); // 342.05
console.log("precise_round(342.055, 2) :", precise_round(342.055, 2)); // 342.06
console.log("precise_round(-342.055, 2) :", precise_round(-342.055, 2)); // -342.06
console.log("inprecise_round(0.565, 2) :", inprecise_round(0.565, 2)); // 0.56
console.log("precise_round(0.565, 2) :", precise_round(0.565, 2)); // 0.57

Here's a simple one
function roundFloat(num,dec){
var d = 1;
for (var i=0; i<dec; i++){
d += "0";
}
return Math.round(num * d) / d;
}
Use like alert(roundFloat(1.79209243929,4));
Jsfiddle

Round down
function round_down(value, decPlaces) {
return Math.floor(value * Math.pow(10, decPlaces)) / Math.pow(10, decPlaces);
}
Round up
function round_up(value, decPlaces) {
return Math.ceil(value * Math.pow(10, decPlaces)) / Math.pow(10, decPlaces);
}
Round nearest
function round_nearest(value, decPlaces) {
return Math.round(value * Math.pow(10, decPlaces)) / Math.pow(10, decPlaces);
}
Merged https://stackoverflow.com/a/7641824/1889449 and
https://www.kirupa.com/html5/rounding_numbers_in_javascript.htm Thanks
them.

Building on top of Christian C. Salvadó's answer, doing the following will output a Number type, and also seems to be dealing with rounding well:
const roundNumberToTwoDecimalPlaces = (num) => Number(new Intl.NumberFormat('en-US', {
minimumFractionDigits: 2,
maximumFractionDigits: 2,
}).format(num));
roundNumberToTwoDecimalPlaces(1.344); // => 1.34
roundNumberToTwoDecimalPlaces(1.345); // => 1.35
The difference between the above and what has already been mentioned is that you don't need the .format() chaining when you're using it[, and that it outputs a Number type].

#heridev and I created a small function in jQuery.
You can try next:
HTML
<input type="text" name="one" class="two-digits"><br>
<input type="text" name="two" class="two-digits">​
jQuery
// apply the two-digits behaviour to elements with 'two-digits' as their class
$( function() {
$('.two-digits').keyup(function(){
if($(this).val().indexOf('.')!=-1){
if($(this).val().split(".")[1].length > 2){
if( isNaN( parseFloat( this.value ) ) ) return;
this.value = parseFloat(this.value).toFixed(2);
}
}
return this; //for chaining
});
});
​
DEMO ONLINE:
http://jsfiddle.net/c4Wqn/

The trouble with floating point values is that they are trying to represent an infinite amount of (continuous) values with a fixed amount of bits. So naturally, there must be some loss in play, and you're going to be bitten with some values.
When a computer stores 1.275 as a floating point value, it won't actually remember whether it was 1.275 or 1.27499999999999993, or even 1.27500000000000002. These values should give different results after rounding to two decimals, but they won't, since for computer they look exactly the same after storing as floating point values, and there's no way to restore the lost data. Any further calculations will only accumulate such imprecision.
So, if precision matters, you have to avoid floating point values from the start. The simplest options are to
use a devoted library
use strings for storing and passing around the values (accompanied by string operations)
use integers (e.g. you could be passing around the amount of hundredths of your actual value, e.g. amount in cents instead of amount in dollars)
For example, when using integers to store the number of hundredths, the function for finding the actual value is quite simple:
function descale(num, decimals) {
var hasMinus = num < 0;
var numString = Math.abs(num).toString();
var precedingZeroes = '';
for (var i = numString.length; i <= decimals; i++) {
precedingZeroes += '0';
}
numString = precedingZeroes + numString;
return (hasMinus ? '-' : '')
+ numString.substr(0, numString.length-decimals)
+ '.'
+ numString.substr(numString.length-decimals);
}
alert(descale(127, 2));
With strings, you'll need rounding, but it's still manageable:
function precise_round(num, decimals) {
var parts = num.split('.');
var hasMinus = parts.length > 0 && parts[0].length > 0 && parts[0].charAt(0) == '-';
var integralPart = parts.length == 0 ? '0' : (hasMinus ? parts[0].substr(1) : parts[0]);
var decimalPart = parts.length > 1 ? parts[1] : '';
if (decimalPart.length > decimals) {
var roundOffNumber = decimalPart.charAt(decimals);
decimalPart = decimalPart.substr(0, decimals);
if ('56789'.indexOf(roundOffNumber) > -1) {
var numbers = integralPart + decimalPart;
var i = numbers.length;
var trailingZeroes = '';
var justOneAndTrailingZeroes = true;
do {
i--;
var roundedNumber = '1234567890'.charAt(parseInt(numbers.charAt(i)));
if (roundedNumber === '0') {
trailingZeroes += '0';
} else {
numbers = numbers.substr(0, i) + roundedNumber + trailingZeroes;
justOneAndTrailingZeroes = false;
break;
}
} while (i > 0);
if (justOneAndTrailingZeroes) {
numbers = '1' + trailingZeroes;
}
integralPart = numbers.substr(0, numbers.length - decimals);
decimalPart = numbers.substr(numbers.length - decimals);
}
} else {
for (var i = decimalPart.length; i < decimals; i++) {
decimalPart += '0';
}
}
return (hasMinus ? '-' : '') + integralPart + (decimals > 0 ? '.' + decimalPart : '');
}
alert(precise_round('1.275', 2));
alert(precise_round('1.27499999999999993', 2));
Note that this function rounds to nearest, ties away from zero, while IEEE 754 recommends rounding to nearest, ties to even as the default behavior for floating point operations. Such modifications are left as an exercise for the reader :)

Round your decimal value, then use toFixed(x) for your expected digit(s).
function parseDecimalRoundAndFixed(num,dec){
var d = Math.pow(10,dec);
return (Math.round(num * d) / d).toFixed(dec);
}
Call
parseDecimalRoundAndFixed(10.800243929,4) => 10.80
parseDecimalRoundAndFixed(10.807243929,2) => 10.81

Number(Math.round(1.005+'e2')+'e-2'); // 1.01
This worked for me: Rounding Decimals in JavaScript

With these examples you will still get an error when trying to round the number 1.005 the solution is to either use a library like Math.js or this function:
function round(value: number, decimals: number) {
return Number(Math.round(value + 'e' + decimals) + 'e-' + decimals);
}

Here is my 1-line solution: Number((yourNumericValueHere).toFixed(2));
Here's what happens:
1) First, you apply .toFixed(2) onto the number that you want to round off the decimal places of. Note that this will convert the value to a string from number. So if you are using Typescript, it will throw an error like this:
"Type 'string' is not assignable to type 'number'"
2) To get back the numeric value or to convert the string to numeric value, simply apply the Number() function on that so-called 'string' value.
For clarification, look at the example below:
EXAMPLE:
I have an amount that has upto 5 digits in the decimal places and I would like to shorten it to upto 2 decimal places. I do it like so:
var price = 0.26453;
var priceRounded = Number((price).toFixed(2));
console.log('Original Price: ' + price);
console.log('Price Rounded: ' + priceRounded);

In general, decimal rounding is done by scaling: round(num * p) / p
Naive implementation
Using the following function with halfway numbers, you will get either the upper rounded value as expected, or the lower rounded value sometimes depending on the input.
This inconsistency in rounding may introduce hard to detect bugs in the client code.
function naiveRound(num, decimalPlaces) {
var p = Math.pow(10, decimalPlaces);
return Math.round(num * p) / p;
}
console.log( naiveRound(1.245, 2) ); // 1.25 correct (rounded as expected)
console.log( naiveRound(1.255, 2) ); // 1.25 incorrect (should be 1.26)
Better implementations
By converting the number to a string in the exponential notation, positive numbers are rounded as expected.
But, be aware that negative numbers round differently than positive numbers.
In fact, it performs what is basically equivalent to "round half up" as the rule, you will see that round(-1.005, 2) evaluates to -1 even though round(1.005, 2) evaluates to 1.01. The lodash _.round method uses this technique.
/**
* Round half up ('round half towards positive infinity')
* Uses exponential notation to avoid floating-point issues.
* Negative numbers round differently than positive numbers.
*/
function round(num, decimalPlaces) {
num = Math.round(num + "e" + decimalPlaces);
return Number(num + "e" + -decimalPlaces);
}
// test rounding of half
console.log( round(0.5, 0) ); // 1
console.log( round(-0.5, 0) ); // 0
// testing edge cases
console.log( round(1.005, 2) ); // 1.01
console.log( round(2.175, 2) ); // 2.18
console.log( round(5.015, 2) ); // 5.02
console.log( round(-1.005, 2) ); // -1
console.log( round(-2.175, 2) ); // -2.17
console.log( round(-5.015, 2) ); // -5.01
If you want the usual behavior when rounding negative numbers, you would need to convert negative numbers to positive before calling Math.round(), and then convert them back to negative numbers before returning.
// Round half away from zero
function round(num, decimalPlaces) {
num = Math.round(Math.abs(num) + "e" + decimalPlaces) * Math.sign(num);
return Number(num + "e" + -decimalPlaces);
}
There is a different purely mathematical technique to perform round-to-nearest (using "round half away from zero"), in which epsilon correction is applied before calling the rounding function.
Simply, we add the smallest possible float value (= 1.0 ulp; unit in the last place) to the number before rounding. This moves to the next representable value after the number, away from zero.
/**
* Round half away from zero ('commercial' rounding)
* Uses correction to offset floating-point inaccuracies.
* Works symmetrically for positive and negative numbers.
*/
function round(num, decimalPlaces) {
var p = Math.pow(10, decimalPlaces);
var e = Number.EPSILON * num * p;
return Math.round((num * p) + e) / p;
}
// test rounding of half
console.log( round(0.5, 0) ); // 1
console.log( round(-0.5, 0) ); // -1
// testing edge cases
console.log( round(1.005, 2) ); // 1.01
console.log( round(2.175, 2) ); // 2.18
console.log( round(5.015, 2) ); // 5.02
console.log( round(-1.005, 2) ); // -1.01
console.log( round(-2.175, 2) ); // -2.18
console.log( round(-5.015, 2) ); // -5.02
This is needed to offset the implicit round-off error that may occur during encoding of decimal numbers, particularly those having "5" in the last decimal position, like 1.005, 2.675 and 16.235. Actually, 1.005 in decimal system is encoded to 1.0049999999999999 in 64-bit binary float; while, 1234567.005 in decimal system is encoded to 1234567.0049999998882413 in 64-bit binary float.
It is worth noting that the maximum binary round-off error is dependent upon (1) the magnitude of the number and (2) the relative machine epsilon (2^-52).

Put the following in some global scope:
Number.prototype.getDecimals = function ( decDigCount ) {
return this.toFixed(decDigCount);
}
and then try:
var a = 56.23232323;
a.getDecimals(2); // will return 56.23
Update
Note that toFixed() can only work for the number of decimals between 0-20 i.e. a.getDecimals(25) may generate a javascript error, so to accomodate that you may add some additional check i.e.
Number.prototype.getDecimals = function ( decDigCount ) {
return ( decDigCount > 20 ) ? this : this.toFixed(decDigCount);
}

Number(((Math.random() * 100) + 1).toFixed(2))
this will return a random number from 1 to 100 rounded to 2 decimal places.

Using this response by reference: https://stackoverflow.com/a/21029698/454827
I build a function to get dynamic numbers of decimals:
function toDec(num, dec)
{
if(typeof dec=='undefined' || dec<0)
dec = 2;
var tmp = dec + 1;
for(var i=1; i<=tmp; i++)
num = num * 10;
num = num / 10;
num = Math.round(num);
for(var i=1; i<=dec; i++)
num = num / 10;
num = num.toFixed(dec);
return num;
}
here working example: https://jsfiddle.net/wpxLduLc/

parse = function (data) {
data = Math.round(data*Math.pow(10,2))/Math.pow(10,2);
if (data != null) {
var lastone = data.toString().split('').pop();
if (lastone != '.') {
data = parseFloat(data);
}
}
return data;
};
$('#result').html(parse(200)); // output 200
$('#result1').html(parse(200.1)); // output 200.1
$('#result2').html(parse(200.10)); // output 200.1
$('#result3').html(parse(200.109)); // output 200.11
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.0.0/jquery.min.js"></script>
<div id="result"></div>
<div id="result1"></div>
<div id="result2"></div>
<div id="result3"></div>

I got some ideas from this post a few months back, but none of the answers here, nor answers from other posts/blogs could handle all the scenarios (e.g. negative numbers and some "lucky numbers" our tester found). In the end, our tester did not find any problem with this method below. Pasting a snippet of my code:
fixPrecision: function (value) {
var me = this,
nan = isNaN(value),
precision = me.decimalPrecision;
if (nan || !value) {
return nan ? '' : value;
} else if (!me.allowDecimals || precision <= 0) {
precision = 0;
}
//[1]
//return parseFloat(Ext.Number.toFixed(parseFloat(value), precision));
precision = precision || 0;
var negMultiplier = value < 0 ? -1 : 1;
//[2]
var numWithExp = parseFloat(value + "e" + precision);
var roundedNum = parseFloat(Math.round(Math.abs(numWithExp)) + 'e-' + precision) * negMultiplier;
return parseFloat(roundedNum.toFixed(precision));
},
I also have code comments (sorry i forgot all the details already)...I'm posting my answer here for future reference:
9.995 * 100 = 999.4999999999999
Whereas 9.995e2 = 999.5
This discrepancy causes Math.round(9.995 * 100) = 999 instead of 1000.
Use e notation instead of multiplying /dividing by Math.Pow(10,precision).

I'm fix the problem the modifier.
Support 2 decimal only.
$(function(){
//input number only.
convertNumberFloatZero(22); // output : 22.00
convertNumberFloatZero(22.5); // output : 22.50
convertNumberFloatZero(22.55); // output : 22.55
convertNumberFloatZero(22.556); // output : 22.56
convertNumberFloatZero(22.555); // output : 22.55
convertNumberFloatZero(22.5541); // output : 22.54
convertNumberFloatZero(22222.5541); // output : 22,222.54
function convertNumberFloatZero(number){
if(!$.isNumeric(number)){
return 'NaN';
}
var numberFloat = number.toFixed(3);
var splitNumber = numberFloat.split(".");
var cNumberFloat = number.toFixed(2);
var cNsplitNumber = cNumberFloat.split(".");
var lastChar = splitNumber[1].substr(splitNumber[1].length - 1);
if(lastChar > 0 && lastChar < 5){
cNsplitNumber[1]--;
}
return Number(splitNumber[0]).toLocaleString('en').concat('.').concat(cNsplitNumber[1]);
};
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>

(Math.round((10.2)*100)/100).toFixed(2)
That should yield: 10.20
(Math.round((.05)*100)/100).toFixed(2)
That should yield: 0.05
(Math.round((4.04)*100)/100).toFixed(2)
That should yield: 4.04
etc.

/*Due to all told stuff. You may do 2 things for different purposes:
When showing/printing stuff use this in your alert/innerHtml= contents:
YourRebelNumber.toFixed(2)*/
var aNumber=9242.16;
var YourRebelNumber=aNumber-9000;
alert(YourRebelNumber);
alert(YourRebelNumber.toFixed(2));
/*and when comparing use:
Number(YourRebelNumber.toFixed(2))*/
if(YourRebelNumber==242.16)alert("Not Rounded");
if(Number(YourRebelNumber.toFixed(2))==242.16)alert("Rounded");
/*Number will behave as you want in that moment. After that, it'll return to its defiance.
*/

This is very simple and works just as well as any of the others:
function parseNumber(val, decimalPlaces) {
if (decimalPlaces == null) decimalPlaces = 0
var ret = Number(val).toFixed(decimalPlaces)
return Number(ret)
}
Since toFixed() can only be called on numbers, and unfortunately returns a string, this does all the parsing for you in both directions. You can pass a string or a number, and you get a number back every time! Calling parseNumber(1.49) will give you 1, and parseNumber(1.49,2) will give you 1.50. Just like the best of 'em!

You could also use the .toPrecision() method and some custom code, and always round up to the nth decimal digit regardless the length of int part.
function glbfrmt (number, decimals, seperator) {
return typeof number !== 'number' ? number : number.toPrecision( number.toString().split(seperator)[0].length + decimals);
}
You could also make it a plugin for a better use.

Here's a TypeScript implementation of https://stackoverflow.com/a/21323330/916734. It also dries things up with functions, and allows for a optional digit offset.
export function round(rawValue: number | string, precision = 0, fractionDigitOffset = 0): number | string {
const value = Number(rawValue);
if (isNaN(value)) return rawValue;
precision = Number(precision);
if (precision % 1 !== 0) return NaN;
let [ stringValue, exponent ] = scientificNotationToParts(value);
let shiftExponent = exponentForPrecision(exponent, precision, Shift.Right);
const enlargedValue = toScientificNotation(stringValue, shiftExponent);
const roundedValue = Math.round(enlargedValue);
[ stringValue, exponent ] = scientificNotationToParts(roundedValue);
const precisionWithOffset = precision + fractionDigitOffset;
shiftExponent = exponentForPrecision(exponent, precisionWithOffset, Shift.Left);
return toScientificNotation(stringValue, shiftExponent);
}
enum Shift {
Left = -1,
Right = 1,
}
function scientificNotationToParts(value: number): Array<string> {
const [ stringValue, exponent ] = value.toString().split('e');
return [ stringValue, exponent ];
}
function exponentForPrecision(exponent: string, precision: number, shift: Shift): number {
precision = shift * precision;
return exponent ? (Number(exponent) + precision) : precision;
}
function toScientificNotation(value: string, exponent: number): number {
return Number(`${value}e${exponent}`);
}

fun Any.twoDecimalPlaces(numInDouble: Double): String {
return "%.2f".format(numInDouble)
}

How to format a float in javascript?

In JavaScript, when converting from a float to a string, how can I get just 2 digits after the decimal point? For example, 0.34 instead of 0.3445434.

There are functions to round numbers. For example:
var x = 5.0364342423;
print(x.toFixed(2));
will print 5.04.
EDIT:
Fiddle

var result = Math.round(original*100)/100;
The specifics, in case the code isn't self-explanatory.
edit: ...or just use toFixed, as proposed by Tim Büthe. Forgot that one, thanks (and an upvote) for reminder :)

Be careful when using toFixed():
First, rounding the number is done using the binary representation of the number, which might lead to unexpected behaviour. For example
(0.595).toFixed(2) === '0.59'
instead of '0.6'.
Second, there's an IE bug with toFixed(). In IE (at least up to version 7, didn't check IE8), the following holds true:
(0.9).toFixed(0) === '0'
It might be a good idea to follow kkyy's suggestion or to use a custom toFixed() function, eg
function toFixed(value, precision) {
var power = Math.pow(10, precision || 0);
return String(Math.round(value * power) / power);
}

One more problem to be aware of, is that toFixed() can produce unnecessary zeros at the end of the number.
For example:
var x=(23-7.37)
x
15.629999999999999
x.toFixed(6)
"15.630000"
The idea is to clean up the output using a RegExp:
function humanize(x){
return x.toFixed(6).replace(/\.?0*$/,'');
}
The RegExp matches the trailing zeros (and optionally the decimal point) to make sure it looks good for integers as well.
humanize(23-7.37)
"15.63"
humanize(1200)
"1200"
humanize(1200.03)
"1200.03"
humanize(3/4)
"0.75"
humanize(4/3)
"1.333333"

var x = 0.3445434
x = Math.round (x*100) / 100 // this will make nice rounding

The key here I guess is to round up correctly first, then you can convert it to String.
function roundOf(n, p) {
const n1 = n * Math.pow(10, p + 1);
const n2 = Math.floor(n1 / 10);
if (n1 >= (n2 * 10 + 5)) {
return (n2 + 1) / Math.pow(10, p);
}
return n2 / Math.pow(10, p);
}
// All edge cases listed in this thread
roundOf(95.345, 2); // 95.35
roundOf(95.344, 2); // 95.34
roundOf(5.0364342423, 2); // 5.04
roundOf(0.595, 2); // 0.60
roundOf(0.335, 2); // 0.34
roundOf(0.345, 2); // 0.35
roundOf(551.175, 2); // 551.18
roundOf(0.3445434, 2); // 0.34
Now you can safely format this value with toFixed(p).
So with your specific case:
roundOf(0.3445434, 2).toFixed(2); // 0.34

There is a problem with all those solutions floating around using multipliers. Both kkyy and Christoph's solutions are wrong unfortunately.
Please test your code for number 551.175 with 2 decimal places - it will round to 551.17 while it should be 551.18 ! But if you test for ex. 451.175 it will be ok - 451.18. So it's difficult to spot this error at a first glance.
The problem is with multiplying: try 551.175 * 100 = 55117.49999999999 (ups!)
So my idea is to treat it with toFixed() before using Math.round();
function roundFix(number, precision)
{
var multi = Math.pow(10, precision);
return Math.round( (number * multi).toFixed(precision + 1) ) / multi;
}

If you want the string without round you can use this RegEx (maybe is not the most efficient way... but is really easy)
(2.34567778).toString().match(/\d+\.\d{2}/)[0]
// '2.34'

function trimNumber(num, len) {
const modulu_one = 1;
const start_numbers_float=2;
var int_part = Math.trunc(num);
var float_part = String(num % modulu_one);
float_part = float_part.slice(start_numbers_float, start_numbers_float+len);
return int_part+'.'+float_part;
}

There is no way to avoid inconsistent rounding for prices with x.xx5 as actual value using either multiplication or division. If you need to calculate correct prices client-side you should keep all amounts in cents. This is due to the nature of the internal representation of numeric values in JavaScript. Notice that Excel suffers from the same problems so most people wouldn't notice the small errors caused by this phenomen. However errors may accumulate whenever you add up a lot of calculated values, there is a whole theory around this involving the order of calculations and other methods to minimize the error in the final result. To emphasize on the problems with decimal values, please note that 0.1 + 0.2 is not exactly equal to 0.3 in JavaScript, while 1 + 2 is equal to 3.

Maybe you'll also want decimal separator? Here is a function I just made:
function formatFloat(num,casasDec,sepDecimal,sepMilhar) {
if (num < 0)
{
num = -num;
sinal = -1;
} else
sinal = 1;
var resposta = "";
var part = "";
if (num != Math.floor(num)) // decimal values present
{
part = Math.round((num-Math.floor(num))*Math.pow(10,casasDec)).toString(); // transforms decimal part into integer (rounded)
while (part.length < casasDec)
part = '0'+part;
if (casasDec > 0)
{
resposta = sepDecimal+part;
num = Math.floor(num);
} else
num = Math.round(num);
} // end of decimal part
while (num > 0) // integer part
{
part = (num - Math.floor(num/1000)*1000).toString(); // part = three less significant digits
num = Math.floor(num/1000);
if (num > 0)
while (part.length < 3) // 123.023.123 if sepMilhar = '.'
part = '0'+part; // 023
resposta = part+resposta;
if (num > 0)
resposta = sepMilhar+resposta;
}
if (sinal < 0)
resposta = '-'+resposta;
return resposta;
}

/** don't spend 5 minutes, use my code **/
function prettyFloat(x,nbDec) {
if (!nbDec) nbDec = 100;
var a = Math.abs(x);
var e = Math.floor(a);
var d = Math.round((a-e)*nbDec); if (d == nbDec) { d=0; e++; }
var signStr = (x<0) ? "-" : " ";
var decStr = d.toString(); var tmp = 10; while(tmp<nbDec && d*tmp < nbDec) {decStr = "0"+decStr; tmp*=10;}
var eStr = e.toString();
return signStr+eStr+"."+decStr;
}
prettyFloat(0); // "0.00"
prettyFloat(-1); // "-1.00"
prettyFloat(-0.999); // "-1.00"
prettyFloat(0.5); // "0.50"

I use this code to format floats. It is based on toPrecision() but it strips unnecessary zeros. I would welcome suggestions for how to simplify the regex.
function round(x, n) {
var exp = Math.pow(10, n);
return Math.floor(x*exp + 0.5)/exp;
}
Usage example:
function test(x, n, d) {
var rounded = rnd(x, d);
var result = rounded.toPrecision(n);
result = result.replace(/\.?0*$/, '');
result = result.replace(/\.?0*e/, 'e');
result = result.replace('e+', 'e');
return result;
}
document.write(test(1.2000e45, 3, 2) + '=' + '1.2e45' + '<br>');
document.write(test(1.2000e+45, 3, 2) + '=' + '1.2e45' + '<br>');
document.write(test(1.2340e45, 3, 2) + '=' + '1.23e45' + '<br>');
document.write(test(1.2350e45, 3, 2) + '=' + '1.24e45' + '<br>');
document.write(test(1.0000, 3, 2) + '=' + '1' + '<br>');
document.write(test(1.0100, 3, 2) + '=' + '1.01' + '<br>');
document.write(test(1.2340, 4, 2) + '=' + '1.23' + '<br>');
document.write(test(1.2350, 4, 2) + '=' + '1.24' + '<br>');

countDecimals = value => {
if (Math.floor(value) === value) return 0;
let stringValue = value.toString().split(".")[1];
if (stringValue) {
return value.toString().split(".")[1].length
? value.toString().split(".")[1].length
: 0;
} else {
return 0;
}
};
formatNumber=(ans)=>{
let decimalPlaces = this.countDecimals(ans);
ans = 1 * ans;
if (decimalPlaces !== 0) {
let onePlusAns = ans + 1;
let decimalOnePlus = this.countDecimals(onePlusAns);
if (decimalOnePlus < decimalPlaces) {
ans = ans.toFixed(decimalPlaces - 1).replace(/\.?0*$/, "");
} else {
let tenMulAns = ans * 10;
let decimalTenMul = this.countDecimals(tenMulAns);
if (decimalTenMul + 1 < decimalPlaces) {
ans = ans.toFixed(decimalPlaces - 1).replace(/\.?0*$/, "");
}
}
}
}
I just add 1 to the value and count the decimal digits present in the original value and the added value. If I find the decimal digits after adding one less than the original decimal digits, I just call the toFixed() with (original decimals - 1). I also check by multiplying the original value by 10 and follow the same logic in case adding one doesn't reduce redundant decimal places.
A simple workaround to handle floating-point number rounding in JS. Works in most cases I tried.

How do you convert numbers between different bases in JavaScript?

I would like to convert numbers between different bases, such as hexadecimal and decimal.
Example: How do you convert hexadecimal 8F to decimal?

The API
To convert to a number from a hex string:
parseInt(string, radix)
string: Required. The string to be parsed
radix: Optional. A number (from 2 to 36) that represents the numeral system to be used
To convert from a number to a hex string:
NumberObject.toString(radix)
radix: Optional. Specifies the base radix you would like the number displayed as.
Example radix values:
2 - The number will show as a binary value
8 - The number will show as an octal value
16 - The number will show as an hexadecimal value
Example Usage
Integer value to hex:
var i = 10;
console.log( i.toString(16) );
Hex string to integer value:
var h = "a";
console.log( parseInt(h, 16) );
Integer value to decimal:
var d = 16;
console.log( d.toString(10) );

Update 2023-01-05: Now supports large numbers and floats via https://github.com/ryansmith94/baseroo
I came to this post needing to convert from base 10 to 62 and vice-versa. Whilst the solutions here are great, parseInt and toString only support base 2 to 36. So if anyone finds themselves in a similar position to me needing base 2 to 62, I've pasted my solution below.
function convertBase(value, from_base, to_base) {
var range = '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ+/'.split('');
var from_range = range.slice(0, from_base);
var to_range = range.slice(0, to_base);
var dec_value = value.split('').reverse().reduce(function (carry, digit, index) {
if (from_range.indexOf(digit) === -1) throw new Error('Invalid digit `'+digit+'` for base '+from_base+'.');
return carry += from_range.indexOf(digit) * (Math.pow(from_base, index));
}, 0);
var new_value = '';
while (dec_value > 0) {
new_value = to_range[dec_value % to_base] + new_value;
dec_value = (dec_value - (dec_value % to_base)) / to_base;
}
return new_value || '0';
}

You may try the following code, which also supports arbitrary precision numbers (larger than 2^53).
function convertBase(str, fromBase, toBase) {
const DIGITS = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ+/";
const add = (x, y, base) => {
let z = [];
const n = Math.max(x.length, y.length);
let carry = 0;
let i = 0;
while (i < n || carry) {
const xi = i < x.length ? x[i] : 0;
const yi = i < y.length ? y[i] : 0;
const zi = carry + xi + yi;
z.push(zi % base);
carry = Math.floor(zi / base);
i++;
}
return z;
}
const multiplyByNumber = (num, x, base) => {
if (num < 0) return null;
if (num == 0) return [];
let result = [];
let power = x;
while (true) {
num & 1 && (result = add(result, power, base));
num = num >> 1;
if (num === 0) break;
power = add(power, power, base);
}
return result;
}
const parseToDigitsArray = (str, base) => {
const digits = str.split('');
let arr = [];
for (let i = digits.length - 1; i >= 0; i--) {
const n = DIGITS.indexOf(digits[i])
if (n == -1) return null;
arr.push(n);
}
return arr;
}
const digits = parseToDigitsArray(str, fromBase);
if (digits === null) return null;
let outArray = [];
let power = [1];
for (let i = 0; i < digits.length; i++) {
digits[i] && (outArray = add(outArray, multiplyByNumber(digits[i], power, toBase), toBase));
power = multiplyByNumber(fromBase, power, toBase);
}
let out = '';
for (let i = outArray.length - 1; i >= 0; i--)
out += DIGITS[outArray[i]];
return out;
}
Usage:
console.log(convertBase("5a2a9c826c75045be9ba8fbffc80c6f25a2a9c826c75045be9ba8fbffc80c6f2",16,64));
// Returns: 5EGD89ItghrWrGfL/O0NL9qaFO2r7k4m+CWzX/YwcrO
console.log(convertBase("5EGD89ItghrWrGfL/O0NL9qaFO2r7k4m+CWzX/YwcrO",64,16));
// Returns: 5a2a9c826c75045be9ba8fbffc80c6f25a2a9c826c75045be9ba8fbffc80c6f2
The basic code has been found here, I have a bit improved it to also support up to base 64.

The following diagram might help. Note that to convert from base 16 to base 2 you need to convert first to base 10 and then base 2.

This function generate decimal number to any base from 2 to 36.(as like javascript)But you can Increase the base more than 36 just by adding new character in keys[] like lowercase["a","b"]
function toBase(num, radix = 10) { // only i64 numbers
var keys = ['🤢', '😃', 2, 3, 4, 5, 6, 7, 8, 9, "A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z"];
if (!(radix >= 2 && radix <= keys.length)) throw new RangeError("toBase() radix argument must be between 2 and " + keys.length)
if (num < 0) var isNegative = true
if (isNaN(num = Math.abs(+num))) return NaN
let output = [];
do {
let index = num % radix;
output.unshift(keys[index]);
num = Math.trunc(num / radix);
} while (num != 0);
if (isNegative) output.unshift('-')
return output.join("");
}
console.log(toBase("100",2))

Specify the radix you want to use as a parameter.
NOTE: This only works to convert from bases 2-36 to decimal and little values.
parseInt(string, radix)
parseInt("80", 10) // results in 80
parseInt("80", 16) // results in 128
// etc
About "little", parseInt("6f", 32) is fine (= 207), but any other little bigger will be also 207, 6f1, 6f11, ...

Well, I made a function that could translate from base 10 to any base. (This depends on how many strings you have in the array A, if it's more than that + 10 it'll run out of symbols), and I almost cried when I found out you could to it in less than 10 characters with that...
Add a bookmark and as URL insert this... I've done it the long but personal way. At least, mine can use a base which is higher than 36. You can add more symbols yourself, but if you want, I can make it for you...
var X = prompt("Choose your number");
var Y = prompt("Choose your base");
var Z = [];
var M = -1;
var A = ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z"];
var B = function() {
for (i = X; 0 < i; i = Math.floor(i / Y)) {
if(i % Y >= 10) {
Z.push(A[i % Y - 10]);
} else {
Z.push(i % Y);
}
M = M + 1;
}
for (j = M; j >= 0; j--) {
document.write(Z[j]);
}
};
B(); // Call function

Usually I use this function to convert from different bases.
For example, it returns "11111111" for both cases:
convertBase("ff", 16, 2) or convertBase(0xFF, 16, 2)
var convertBase = function(val, base1, base2) {
if (typeof(val) == "number") {
return parseInt(String(val)).toString(base2);
} else {
return parseInt(val.toString(), base1).toString(base2)
};
}

I've written a function to convert a JavaScript string from one base to another base, with the original base and the new base specified as parameters.
function convertFromBaseToBase(str, fromBase, toBase){
var num = parseInt(str, fromBase);
return num.toString(toBase);
}
alert(convertFromBaseToBase(10, 2, 10));

This function converts a number from base 10, to an arbitrary base:
function to_base(base, num) {
const largest_power = ~~(Math.log(num) / Math.log(base));
const result = [];
for (let pow = largest_power; pow >= 0; pow--) {
const digit = ~~(num / base ** pow);
num -= digit * base ** pow;
result.push(digit);
}
return result;
}
to_base(2, 13) // [1, 1, 0, 1]
to_base(10, 458) // [4, 5, 8]
to_base(32, 1024) // [1, 0, 0]
to_base(32, 1023) // [31, 31]
which can be useful if you want to use weird bases along with weird char-set

Using the parseInt function:
var noInBase10 = parseInt('8F',16);

Check the complete JS code to convert into different base
/**
* Convert From/To Binary/Decimal/Hexadecimal in JavaScript
* https://gist.github.com/shamshul2007/
* Copyright 2012-2015, Shamshul <shamshul2007#gmail.com>
* Licensed under The MIT License
* http://www.opensource.org/licenses/mit-license
*/
(function(){
var ConvertBase = function (num) {
return {
from : function (baseFrom) {
return {
to : function (baseTo) {
return parseInt(num, baseFrom).toString(baseTo);
}
};
}
};
};
// binary to decimal
ConvertBase.bin2dec = function (num) {
return ConvertBase(num).from(2).to(10);
};
// binary to hexadecimal
ConvertBase.bin2hex = function (num) {
return ConvertBase(num).from(2).to(16);
};
// decimal to binary
ConvertBase.dec2bin = function (num) {
return ConvertBase(num).from(10).to(2);
};
// decimal to hexadecimal
ConvertBase.dec2hex = function (num) {
return ConvertBase(num).from(10).to(16);
};
// hexadecimal to binary
ConvertBase.hex2bin = function (num) {
return ConvertBase(num).from(16).to(2);
};
// hexadecimal to decimal
ConvertBase.hex2dec = function (num) {
return ConvertBase(num).from(16).to(10);
};
//Octal to Decimal
ConvertBase.oct2dec = function (num) {
return ConvertBase(num).from(8).to(10);
};
//Decimal to Octal
ConvertBase.dec2oct = function (num) {
return ConvertBase(num).from(10).to(8);
};
this.ConvertBase = ConvertBase;
})(this);
/*
* Usage example:
* ConvertBase.bin2dec('1111'); // '15'
* ConvertBase.dec2hex('82'); // '52'
* ConvertBase.hex2bin('e2'); // '11100010'
* ConvertBase.dec2bin('153'); // '10011001'
* ConvertBase.hex2dec('1FE4ED63D55FA51E'); //'2298222722903156000'
* ConvertBase.oct2dec('777'); //'511'
* ConvertBase.dec2oct('551'); //'1047'
*/

Try the following code, optimised from Slavik Meltser's post, implements BASE n conversions with all the radix combinations between Base2 and Base256. This code accepts three sorts of arguments to define source and destination number systems:
by number system radix (e.g. 8)
by number system convention name (e.g. 'Bitcoin')
by giving custom numerals as an argument (e.g. ['0123456789ABCDEF'])
You'll see that some number system numerals have been hard-coded inside the class and will be used as default when you pass the radix as an argument. (.e.g. 64) When no hard-coded numeral exists (e.g. 16), default numerals are assigned for all radixes ranging between Base2 and Base256, which becomes very clear in the self-test further below.
function BASE() {
/**
* BASE n converter doing all the radix combinations between Base2 and Base256
* #param {String} str input number
* #param {Number|String|Array} fromBase input number system radix (Number, e.g. 64), convention name (String, e.g. 'Bitcoin') or range (Array)
* #param {Number|String|Array} toBASE output number system radix (Number), convention name (String) or range (Array e.g. ['0123456789'])
* #return {String} output number
*/
this.convert = function (str, fromBase, toBASE)
{
if(typeof(fromBase)=='object') { this.fromSymbols = fromBase[0] } else this.fromSymbols = this.getsymbols(fromBase);
if(typeof(toBASE) =='object') { this.toSymbols = toBASE[0] } else this.toSymbols = this.getsymbols(toBASE);
fromBase = this.fromSymbols.length; toBASE = this.toSymbols.length;
// PARSE INPUT DIGITS ARRAY
for(var _a = [0], str = str.split(''); str.length > 0 && _a[_a.push(this.fromSymbols.indexOf(str.pop())) - 1] >= 0;);
var _d = _a.shift() + _a[_a.length-1]>=0 ? _a : null; if (_d === null) return null;
// BASE CONVERSION
for (var _n = 0,_a = [],_p = [1]; _n < _d.length; _n++) { _a = add(_a, mul(_d[_n], _p, toBASE), toBASE); _p = mul(fromBase, _p, toBASE) }
// PARSE OUTPUT DIGITS ARRAY
for (var _n = _a.length - 1, _o = ''; _n >= 0; _o += this.toSymbols[_a[_n--]]);
return _o.length==0?this.toSymbols[0]:_o;
}
this.symbols = {
32:function(){return this["base32hex"]},
36:["[0-9][A-Z]"],
45:function(){return this["qr-alnum"]},
58:function(){return this["Bitcoin"]},
64:["[A-Z][a-z][0-9]+/"],
85:function(){return this["RFC 1924"]},
91:["[A-Z][a-z][0-9]!#$%&()*+,./:;<=>?#[]^_`{|}~\""],
94:["[!-~]"],
"geohash": ["[0-9][b-h]jkmn[p-z]"], // base 32
"RFC 4648": ["[A-Z][2-7]"], // base 32
"base32hex": ["[0-9][A-V]"], // base 32
"qr-alnum":["[0-9][A-Z] $%*+-./:"], // base 45
"Bitcoin": ["[1-9][A-H]JKLMN[P-Z][a-k][m-z]"], // base 58
"RFC 1924": ["[0-9][A-Z][a-z]!#$%&()*+-;<=>?#^_`{|}~"] // base 85
}
this.getsymbols = function(index) {
if(typeof(this.symbols[index])=="undefined") this.symbols[index] = index<95?this.rng(index<64?"[0-9][A-Z][a-z]+":"[A-Z][a-z][0-9][!-/][:-#][[-`][{-~]").substring(0,index):this.rng("[\x00-\xff]").substring(256-index,256);
if(typeof(this.symbols[index])=="function") this.symbols[index] = this.symbols[index](); // process references
if(typeof(this.symbols[index])=="object") this.symbols[index] = this.rng(this.symbols[index][0]); // process range_replace
return this.symbols[index];
}
this.rng = function(_s) {
var _a = _s.match(/\[.-.\]/); if(_a==null) return _s; else { _a=[_a[0].charCodeAt(1),_a[0].charCodeAt(3)];
return this.rng(_s.replace(RegExp("\\[(\\x"+("0"+_a[0].toString(16)).slice(-2)+"-\\x"+_a[1].toString(16)+")\\]","g")
,String.fromCharCode(..." ".repeat(_a[1]-_a[0]+1).split("").map((_e,_i)=>_i+_a[0])) )) }
}
this.selftest = function() {
var _a={}; for(var _o in this.symbols) _a[_o] = this.getsymbols(_o).length; // built-in symbols
for(_o=2;_o<=95;_o++) _a[_o] = this.getsymbols(_o).length; _a[256]=256; // symbol range 2-95 + 256 (96-255 is similar)
var _s = "",_a = Object.keys(_a).sort(function(a,b){return _a[a]-_a[b]}); // sort merged list
for(var _i in _a) { // iterate number systems
_o = {fromBase:10, toBASE:_a[_i]}; var _r = this.convert("",10,_o.toBASE)
_s += "\r\n\oBASE.convert(n, '"+_o.fromBase+"', '"+_o.toBASE+"') ["+this.fromSymbols+"] ["+this.toSymbols+"]\r\n"
for(var _n=0;_n<(this.fromSymbols.length+2);_n++) { // iterate numbers
_r = this.convert(String(_n),_o.fromBase,_o.toBASE)
_s += _n+(String(_n)==this.convert(_r,_o.toBASE,_o.fromBase)?">":"?")+"["+_r+"] ";
}
}
return _s
}
var add = function(x, y, base) {
var _m = Math.max(x.length, y.length);
for(var _c = _n = 0,_r = []; _n < _m || _c; _c = Math.floor(_z / base)) {
var _z = _c + (_n < x.length ? x[_n] : 0) + (_n < y.length ? y[_n] : 0);
var _n = _r.push(_z % base);
}
return _r;
}
var mul = function(x, pow, base) {
for(var _r = x < 0 ? null : []; x > 0; x = x >> 1) {
if(x & 1) _r = add(_r, pow, base);
pow = add(pow, pow, base);
}
return _r;
}
}
Usage:
// quick test, convert from base45 to base32, using custom symbols for base85 and back to base45
var oBASE = new BASE();
var n = "THIS IS A NUMBER"; // Base 45 code = 'qr-alnum'
console.log(n); // Result: 'THIS IS A NUMBER'
var n = oBASE.convert(n,"qr-alnum",32); // Base 45 to Base 32 = 'base32hex'
console.log(n); // Result: '4ONI84LCTLJ1U08G1N'
var s85 = oBASE.rng("[0-9][a-z][A-Z].-:+=^!/*?&<>()[]{}#%$#"); // 32/Z85 custom symbols
var n = oBASE.convert(n,"base32hex",[s85]); // 'base2hex' to custom Base 85
console.log(n); // Result: 'fnaxrZP)?5d[DG'
var n = oBASE.convert(n,[s85],45); // Custom Base 85 to Base 45 = 'qr-alnum'
console.log(n); // Result: 'THIS IS A NUMBER'
function BASE(){this.convert=function(o,r,n){this.fromSymbols="object"==typeof r?r[0]:this.getsymbols(r),this.toSymbols="object"==typeof n?n[0]:this.getsymbols(n),r=this.fromSymbols.length,n=this.toSymbols.length;var i=[0];for(o=o.split("");o.length>0&&i[i.push(this.fromSymbols.indexOf(o.pop()))-1]>=0;);var h=i.shift()+i[i.length-1]>=0?i:null;if(null===h)return null;for(var e=0,l=(i=[],[1]);e<h.length;e++)i=t(i,s(h[e],l,n),n),l=s(r,l,n);e=i.length-1;for(var m="";e>=0;m+=this.toSymbols[i[e--]]);return 0==m.length?this.toSymbols[0]:m},this.symbols={32:function(){return this.base32hex},36:["[0-9][A-Z]"],45:function(){return this["qr-alnum"]},58:function(){return this.Bitcoin},64:["[A-Z][a-z][0-9]+/"],85:function(){return this["RFC 1924"]},91:['[A-Z][a-z][0-9]!#$%&()*+,./:;<=>?#[]^_`{|}~"'],94:["[!-~]"],geohash:["[0-9][b-h]jkmn[p-z]"],"RFC 4648":["[A-Z][2-7]"],base32hex:["[0-9][A-V]"],"qr-alnum":["[0-9][A-Z] $%*+-./:"],Bitcoin:["[1-9][A-H]JKLMN[P-Z][a-k][m-z]"],"RFC 1924":["[0-9][A-Z][a-z]!#$%&()*+-;<=>?#^_`{|}~"]},this.getsymbols=function(t){return void 0===this.symbols[t]&&(this.symbols[t]=t<95?this.rng(t<64?"[0-9][A-Z][a-z]+":"[A-Z][a-z][0-9][!-/][:-#][[-`][{-~]").substring(0,t):this.rng("[\0-ÿ]").substring(256-t,256)),"function"==typeof this.symbols[t]&&(this.symbols[t]=this.symbols[t]()),"object"==typeof this.symbols[t]&&(this.symbols[t]=this.rng(this.symbols[t][0])),this.symbols[t]},this.rng=function(t){var s=t.match(/\[.-.\]/);return null==s?t:(s=[s[0].charCodeAt(1),s[0].charCodeAt(3)],this.rng(t.replace(RegExp("\\[(\\x"+("0"+s[0].toString(16)).slice(-2)+"-\\x"+s[1].toString(16)+")\\]","g"),String.fromCharCode(..." ".repeat(s[1]-s[0]+1).split("").map((t,o)=>o+s[0])))))},this.selftest=function(){var t={};for(var s in this.symbols)t[s]=this.getsymbols(s).length;for(s=2;s<=95;s++)t[s]=this.getsymbols(s).length;t[256]=256;var o="";t=Object.keys(t).sort(function(s,o){return t[s]-t[o]});for(var r in t){s={fromBase:10,toBASE:t[r]};var n=this.convert("",10,s.toBASE);o+="\r\noBASE.convert(n, '"+s.fromBase+"', '"+s.toBASE+"') ["+this.fromSymbols+"] ["+this.toSymbols+"]\r\n";for(var i=0;i<this.fromSymbols.length+2;i++)n=this.convert(String(i),s.fromBase,s.toBASE),o+=i+(String(i)==this.convert(n,s.toBASE,s.fromBase)?">":"?")+"["+n+"] "}return o};var t=function(t,s,o){for(var r=Math.max(t.length,s.length),n=e=0,i=[];e<r||n;n=Math.floor(h/o))var h=n+(e<t.length?t[e]:0)+(e<s.length?s[e]:0),e=i.push(h%o);return i},s=function(s,o,r){for(var n=s<0?null:[];s>0;s>>=1)1&s&&(n=t(n,o,r)),o=t(o,o,r);return n}}
Self-Test:
// quick test, convert from base45 to base32, using custom symbols for base85 and back to base45
var oBASE = new BASE();
console.log(oBASE.selftest())
function BASE(){this.convert=function(o,r,n){this.fromSymbols="object"==typeof r?r[0]:this.getsymbols(r),this.toSymbols="object"==typeof n?n[0]:this.getsymbols(n),r=this.fromSymbols.length,n=this.toSymbols.length;var i=[0];for(o=o.split("");o.length>0&&i[i.push(this.fromSymbols.indexOf(o.pop()))-1]>=0;);var h=i.shift()+i[i.length-1]>=0?i:null;if(null===h)return null;for(var e=0,l=(i=[],[1]);e<h.length;e++)i=t(i,s(h[e],l,n),n),l=s(r,l,n);e=i.length-1;for(var m="";e>=0;m+=this.toSymbols[i[e--]]);return 0==m.length?this.toSymbols[0]:m},this.symbols={32:function(){return this.base32hex},36:["[0-9][A-Z]"],45:function(){return this["qr-alnum"]},58:function(){return this.Bitcoin},64:["[A-Z][a-z][0-9]+/"],85:function(){return this["RFC 1924"]},91:['[A-Z][a-z][0-9]!#$%&()*+,./:;<=>?#[]^_`{|}~"'],94:["[!-~]"],geohash:["[0-9][b-h]jkmn[p-z]"],"RFC 4648":["[A-Z][2-7]"],base32hex:["[0-9][A-V]"],"qr-alnum":["[0-9][A-Z] $%*+-./:"],Bitcoin:["[1-9][A-H]JKLMN[P-Z][a-k][m-z]"],"RFC 1924":["[0-9][A-Z][a-z]!#$%&()*+-;<=>?#^_`{|}~"]},this.getsymbols=function(t){return void 0===this.symbols[t]&&(this.symbols[t]=t<95?this.rng(t<64?"[0-9][A-Z][a-z]+":"[A-Z][a-z][0-9][!-/][:-#][[-`][{-~]").substring(0,t):this.rng("[\0-ÿ]").substring(256-t,256)),"function"==typeof this.symbols[t]&&(this.symbols[t]=this.symbols[t]()),"object"==typeof this.symbols[t]&&(this.symbols[t]=this.rng(this.symbols[t][0])),this.symbols[t]},this.rng=function(t){var s=t.match(/\[.-.\]/);return null==s?t:(s=[s[0].charCodeAt(1),s[0].charCodeAt(3)],this.rng(t.replace(RegExp("\\[(\\x"+("0"+s[0].toString(16)).slice(-2)+"-\\x"+s[1].toString(16)+")\\]","g"),String.fromCharCode(..." ".repeat(s[1]-s[0]+1).split("").map((t,o)=>o+s[0])))))},this.selftest=function(){var t={};for(var s in this.symbols)t[s]=this.getsymbols(s).length;for(s=2;s<=95;s++)t[s]=this.getsymbols(s).length;t[256]=256;var o="";t=Object.keys(t).sort(function(s,o){return t[s]-t[o]});for(var r in t){s={fromBase:10,toBASE:t[r]};var n=this.convert("",10,s.toBASE);o+="\r\noBASE.convert(n, '"+s.fromBase+"', '"+s.toBASE+"') ["+this.fromSymbols+"] ["+this.toSymbols+"]\r\n";for(var i=0;i<this.fromSymbols.length+2;i++)n=this.convert(String(i),s.fromBase,s.toBASE),o+=i+(String(i)==this.convert(n,s.toBASE,s.fromBase)?">":"?")+"["+n+"] "}return o};var t=function(t,s,o){for(var r=Math.max(t.length,s.length),n=e=0,i=[];e<r||n;n=Math.floor(h/o))var h=n+(e<t.length?t[e]:0)+(e<s.length?s[e]:0),e=i.push(h%o);return i},s=function(s,o,r){for(var n=s<0?null:[];s>0;s>>=1)1&s&&(n=t(n,o,r)),o=t(o,o,r);return n}}

to convert numbers into different bases in JavaScript or typescript using the following ways.
function binary(number) {
console.log((number >>> 0).toString(2)); //base 2 for binary
}
binary(-7);
function octal(number) {
console.log(number.toString(8)); //base 8 for octal
}
octal(15);
function hex(number) {
console.log(number.toString(16)); //base 16 for hex
}
hex(15);
here in the binary function, you can use number.toString(2) function, but the problem occurs when representing negative numbers. so that you can use the unsigned right shift bitwise operator (>>>) to fix this issue.

I am writing this answer simply because I was kinda stupid to understand the difference between parseInt and .toString().
Here is the difference
parseInt
parseInt(str, base) will convert the str to an Integer and base here is used to tell parseInt which base the str is in. I thought the base here was what the str would get converted to. But thats not true. parseInt is just used to convert a number in any base to an integer in base10.
.toString()
number.toString(base) is used to convert the number into a number with base base and it assumes that number is in Integer form. So if the number is in hex format and you want to convert it to binary you need to convert the number first to Integer using parseInt and then you can use this function to change the base.
Hope this helps if you are as stupid as me. :)

You can also convert number in hexadecimal to decimal as follows:
var a="8F";
var b=a.split("");
var result=0;var hex_multiplier=1;
for(var i=0;i<b.length;i++){
result +=parseInt(b[i],16)*hex_multiplier;
hex_multiplier *=16;
}
console.log(result);
where you can change a with any hexadecimal number and get the result in decimal form.

You can use JavaScript's built-in integer literals for some scenarios:
function binaryToDecimal(binaryString) {
return Number('0b' + binaryString.replace('-', '')) * signOf(binaryString);;
}
function octalToDecimal(octalString) {
return Number('0o' + octalString.replace('-', '')) * signOf(octalString);
}
function hexToDecimal(hexString) {
return Number('0x' + hexString.replace('-', '')) * signOf(hexString);
}
function signOf(n) {
return n.trim()[0] == '-' ? -1 : 1;
}
console.log(binaryToDecimal('-0101'),
octalToDecimal('-052171'),
hexToDecimal('deadbeef'));

This code converts a number to an arbitrary-large base, outputting an array of digits.
function convertBase(num, base) {
var result = [];
while(num >= 1) {
result.unshift(num % base);
num = Math.floor(num / base);
}
return result;
}
console.log(convertBase(100, 12)); // [8, 4]
console.log(convertBase(5114, 64)); // [1, 15, 58]

TypeScript Implementation that supports decimal and int:
class Digit {
private readonly digitMap: Map<string, number>;
constructor(readonly digits: string, readonly maxDigits = 10) {
this.digitMap = new Map<string, number>();
for (let i = 0; i < digits.length; i += 1) {
if (digits[i] === ".") {
throw new TypeError(`Digits should not contain the "." mark.`);
}
if (this.digitMap.has(digits[i])) {
throw new TypeError(`Duplicated digit character.`);
}
this.digitMap.set(digits[i], i);
}
}
to = (x: number) => {
const int = Math.floor(x);
const dec = x - int;
let intRemains = int;
let intResult = "";
while (true) {
intResult = this.digits[intRemains % this.digits.length] + intResult;
intRemains = Math.floor(intRemains / this.digits.length);
if (intRemains === 0) break;
}
if (dec > 0) {
let decRemains = dec;
let decResult = "";
for (let i = 0; i < this.maxDigits; i += 1) {
const stepCache = decRemains * this.digits.length;
const decInt = Math.floor(stepCache);
decResult = decResult + this.digits[decInt];
decRemains = stepCache - decInt;
if (decRemains === 0) break;
}
if (decResult === "0") return intResult;
return intResult + "." + decResult;
}
return intResult;
};
from = (x: string) => {
const splitted = x.split(".");
if (splitted.length > 2) return Number.NaN;
const [int, dec] = splitted;
let result = 0;
for (let i = 0; i < int.length; i += 1) {
const digit = int[int.length - i - 1];
const oct = this.digitMap.get(digit);
if (oct === undefined) return Number.NaN;
result += oct * Math.pow(this.digits.length, i);
}
if (dec) {
for (let i = 0; i < dec.length; i += 1) {
const digit = dec[i];
const oct = this.digitMap.get(digit);
if (oct === undefined) return Number.NaN;
result += oct * Math.pow(this.digits.length, 0 - i - 1);
}
}
return result;
};
}
Here're the example about how to use it:
Creating the converter:
const d = new Digit("0123456789");
Convert a number to string:
d.to(1.2345);
Convert a string to number:
d.from("1.2345");
You can also set the second parameter to control the maximum number of decimal places, for example:
const d0 = new Digit("abcdABCD", 10);
d0.to(0.96);
// a.DBdAbcbDcD
const d1 = new Digit("abcdABCD", 20);
d1.to(0.96);
// a.DBdAbcbDcDacAdCBC

Monkey patched Number prototype for all radixes up to and including 64
( inspired by and derived from #nirvana's answer )
NumberToStringPatched.__patched = NumberToStringPatched;
if (Number.prototype.toString.__patched!==NumberToStringPatched) {
NumberToStringPatched.__unpatched = Number.prototype.toString;
Number.prototype.toString = NumberToStringPatched;
}
NumberParseIntPatched.__patched = NumberParseIntPatched;
if (Number.parseInt.__patched!==NumberParseIntPatched) {
NumberParseIntPatched.__unpatched = Number.parseInt;
Number.parseInt = NumberParseIntPatched;
}
function NumberToStringPatched(base=10) {
if (base<33||base>64) return NumberToStringPatched.__unpatched.call(this,base);
return convertToBase(this,base);
}
function NumberParseIntPatched(str,base=10) {
if (base<33||base>64) return NumberParseIntPatched.__unpatched.call(this,str,base);
return convertFromBase(str,base);
}
const numberCharSet=(()=>{
let chars = "0123456789";
for (let c='a';c<='z';c=String.fromCharCode(c.charCodeAt(0)+1)) {
chars+=c;
}
for (let c='A';c<='Z';c=String.fromCharCode(c.charCodeAt(0)+1)) {
chars+=c;
}
return (chars+'_$').split('');
})();
function convertToBase(num, base) {
var result = '';
while(num >= 1) {
result = numberCharSet[num % base]+result;
num = Math.floor(num / base);
}
return result;
}
function convertFromBase(str, base) {
let charset = numberCharSet.slice(0,base);
if (typeof str==='number') str=str.toString(base);
if (base <= 26) str=str.toLowerCase();
let digits = str.split('').map(function(c){
let x = charset.indexOf(c);
if (x<0) throw new Error("invalid digit for given radix "+base+': '+c);
return x;
});
let result = digits.shift();
while (digits.length) {
result=(result*base)+digits.shift();
}
return result;
}
for (let i = 0;i<64;i++) {
console.log(i.toString(64));
}
console.log(9000);
console.log(9 * 10 * 10 * 10);
console.log(0xf000);
console.log(15 * 16 * 16 * 16);
console.log(Number.parseInt('$000',64));
console.log(63 * 64 * 64 * 64);

Your Own recursion method for creating base2 conversions.
As mentioned by the users above
let n = 13; console.log(n.toString(2)); will result in 13 conversion from base 10 to base 2.
But in case if you want to program the same. I have written a recursive method to do the same. which just simply divide by 2 and then count remainders.
// #author Tarandeep Singh :: Created recursive converter from base 10 to base 2
// #date : 2017-04-11
// Convert Base 10 to Base 2, We should reverse the output
// For Example base10to2(10) = "0101" just do res = base10to2(10).split('').reverse().join();
function base10to2(val, res = '') {
if (val >= 2) {
res += '' + val % 2;
return base10to2(val = Math.floor(val / 2), res);
} else {
res += '' + 1
return res;
}
}
let n = 13;
var result = base10to2(n).split('').reverse().join();
document.write(`Converting ${n} into Base2 is ${result}`);