Related
I am in need of a JavaScript function which can take a value and pad it to a given length (I need spaces, but anything would do). I found this, but I have no idea what the heck it is doing and it doesn't seem to work for me.
String.prototype.pad = function(l, s, t) {
return s || (s = " "),
(l -= this.length) > 0 ?
(s = new Array(Math.ceil(l / s.length) + 1).join(s))
.substr(0, t = !t ? l : t == 1 ?
0 :
Math.ceil(l / 2)) + this + s.substr(0, l - t) :
this;
};
var s = "Jonas";
document.write(
'<h2>S = '.bold(), s, "</h2>",
'S.pad(20, "[]", 0) = '.bold(), s.pad(20, "[]", 0), "<br />",
'S.pad(20, "[====]", 1) = '.bold(), s.pad(20, "[====]", 1), "<br />",
'S.pad(20, "~", 2) = '.bold(), s.pad(20, "~", 2)
);
ECMAScript 2017 (ES8) added String.padStart (along with String.padEnd) for just this purpose:
"Jonas".padStart(10); // Default pad string is a space
"42".padStart(6, "0"); // Pad with "0"
"*".padStart(8, "-/|\\"); // produces '-/|\\-/|*'
If not present in the JavaScript host, String.padStart can be added as a polyfill.
Pre ES8
I found this solution here and this is for me much much simpler:
var n = 123
String("00000" + n).slice(-5); // returns 00123
("00000" + n).slice(-5); // returns 00123
(" " + n).slice(-5); // returns " 123" (with two spaces)
And here I made an extension to the string object:
String.prototype.paddingLeft = function (paddingValue) {
return String(paddingValue + this).slice(-paddingValue.length);
};
An example to use it:
function getFormattedTime(date) {
var hours = date.getHours();
var minutes = date.getMinutes();
hours = hours.toString().paddingLeft("00");
minutes = minutes.toString().paddingLeft("00");
return "{0}:{1}".format(hours, minutes);
};
String.prototype.format = function () {
var args = arguments;
return this.replace(/{(\d+)}/g, function (match, number) {
return typeof args[number] != 'undefined' ? args[number] : match;
});
};
This will return a time in the format "15:30".
A faster method
If you are doing this repeatedly, for example to pad values in an array, and performance is a factor, the following approach can give you nearly a 100x advantage in speed (jsPerf) over other solution that are currently discussed on the inter webs. The basic idea is that you are providing the pad function with a fully padded empty string to use as a buffer. The pad function just appends to string to be added to this pre-padded string (one string concat) and then slices or trims the result to the desired length.
function pad(pad, str, padLeft) {
if (typeof str === 'undefined')
return pad;
if (padLeft) {
return (pad + str).slice(-pad.length);
} else {
return (str + pad).substring(0, pad.length);
}
}
For example, to zero pad a number to a length of 10 digits,
pad('0000000000',123,true);
To pad a string with whitespace, so the entire string is 255 characters,
var padding = Array(256).join(' '), // make a string of 255 spaces
pad(padding,123,true);
Performance Test
See the jsPerf test here.
And this is faster than ES6 string.repeat by 2x as well, as shown by the revised JsPerf here
Please note that jsPerf is no longer online
Please note that the jsPerf site that we originally used to benchmark the various methods is no longer online. Unfortunately, this means we can't get to those test results. Sad but true.
String.prototype.padStart() and String.prototype.padEnd() are currently TC39 candidate proposals: see github.com/tc39/proposal-string-pad-start-end (only available in Firefox as of April 2016; a polyfill is available).
http://www.webtoolkit.info/javascript_pad.html
/**
*
* JavaScript string pad
* http://www.webtoolkit.info/
*
**/
var STR_PAD_LEFT = 1;
var STR_PAD_RIGHT = 2;
var STR_PAD_BOTH = 3;
function pad(str, len, pad, dir) {
if (typeof(len) == "undefined") { var len = 0; }
if (typeof(pad) == "undefined") { var pad = ' '; }
if (typeof(dir) == "undefined") { var dir = STR_PAD_RIGHT; }
if (len + 1 >= str.length) {
switch (dir){
case STR_PAD_LEFT:
str = Array(len + 1 - str.length).join(pad) + str;
break;
case STR_PAD_BOTH:
var padlen = len - str.length;
var right = Math.ceil( padlen / 2 );
var left = padlen - right;
str = Array(left+1).join(pad) + str + Array(right+1).join(pad);
break;
default:
str = str + Array(len + 1 - str.length).join(pad);
break;
} // switch
}
return str;
}
It's a lot more readable.
Here's a recursive approach to it.
function pad(width, string, padding) {
return (width <= string.length) ? string : pad(width, padding + string, padding)
}
An example...
pad(5, 'hi', '0')
=> "000hi"
ECMAScript 2017 adds a padStart method to the String prototype. This method will pad a string with spaces to a given length. This method also takes an optional string that will be used instead of spaces for padding.
'abc'.padStart(10); // " abc"
'abc'.padStart(10, "foo"); // "foofoofabc"
'abc'.padStart(6,"123465"); // "123abc"
'abc'.padStart(8, "0"); // "00000abc"
'abc'.padStart(1); // "abc"
A padEnd method was also added that works in the same manner.
For browser compatibility (and a useful polyfill) see this link.
Using the ECMAScript 6 method String#repeat, a pad function is as simple as:
String.prototype.padLeft = function(char, length) {
return char.repeat(Math.max(0, length - this.length)) + this;
}
String#repeat is currently supported in Firefox and Chrome only. for other implementation, one might consider the following simple polyfill:
String.prototype.repeat = String.prototype.repeat || function(n){
return n<=1 ? this : (this + this.repeat(n-1));
}
Using the ECMAScript 6 method String#repeat and Arrow functions, a pad function is as simple as:
var leftPad = (s, c, n) => c.repeat(n - s.length) + s;
leftPad("foo", "0", 5); //returns "00foo"
jsfiddle
edit:
suggestion from the comments:
const leftPad = (s, c, n) => n - s.length > 0 ? c.repeat(n - s.length) + s : s;
this way, it wont throw an error when s.lengthis greater than n
edit2:
suggestion from the comments:
const leftPad = (s, c, n) =>{ s = s.toString(); c = c.toString(); return s.length > n ? s : c.repeat(n - s.length) + s; }
this way, you can use the function for strings and non-strings alike.
The key trick in both those solutions is to create an array instance with a given size (one more than the desired length), and then to immediately call the join() method to make a string. The join() method is passed the padding string (spaces probably). Since the array is empty, the empty cells will be rendered as empty strings during the process of joining the array into one result string, and only the padding will remain. It's a really nice technique.
With ES8, there are two options for padding.
You can check them in the documentation.
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padEnd
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padStart
Taking up Samuel's ideas, upward here. And remember an old SQL script, I tried with this:
a=1234;
'0000'.slice(a.toString().length)+a;
It works in all the cases I could imagine:
a= 1 result 0001
a= 12 result 0012
a= 123 result 0123
a= 1234 result 1234
a= 12345 result 12345
a= '12' result 0012
Pad with default values
I noticed that I mostly need the padLeft for time conversion / number padding.
So I wrote this function:
function padL(a, b, c) { // string/number, length=2, char=0
return (new Array(b || 2).join(c || 0) + a).slice(-b)
}
This simple function supports Number or String as input.
The default pad is two characters.
The default char is 0.
So I can simply write:
padL(1);
// 01
If I add the second argument (pad width):
padL(1, 3);
// 001
The third parameter (pad character)
padL('zzz', 10, 'x');
// xxxxxxxzzz
#BananaAcid: If you pass a undefined value or a 0 length string, you get 0undefined, so:
As suggested
function padL(a, b, c) { // string/number, length=2, char=0
return (new Array((b || 1) + 1).join(c || 0) + (a || '')).slice(-(b || 2))
}
But this can also be achieved in a shorter way.
function padL(a, b, c) { // string/number, length=2, char=0
return (new Array(b || 2).join(c || 0) + (a || c || 0)).slice(-b)
}
It also works with:
padL(0)
padL(NaN)
padL('')
padL(undefined)
padL(false)
And if you want to be able to pad in both ways:
function pad(a, b, c, d) { // string/number, length=2, char=0, 0/false=Left-1/true=Right
return a = (a || c || 0), c = new Array(b || 2).join(c || 0), d ? (a + c).slice(0, b) : (c + a).slice(-b)
}
which can be written in a shorter way without using slice.
function pad(a, b, c, d) {
return a = (a || c || 0) + '', b = new Array((++b || 3) - a.length).join(c || 0), d ? a+b : b+a
}
/*
Usage:
pad(
input // (int or string) or undefined, NaN, false, empty string
// default:0 or PadCharacter
// Optional
,PadLength // (int) default:2
,PadCharacter // (string or int) default:'0'
,PadDirection // (bolean) default:0 (padLeft) - (true or 1) is padRight
)
*/
Now if you try to pad 'averylongword' with 2... that’s not my problem.
I said that I would give you a tip.
Most of the time, if you pad, you do it for the same value N times.
Using any type of function inside a loop slows down the loop!!!
So if you just want to pad left some numbers inside a long list, don't use functions to do this simple thing.
Use something like this:
var arrayOfNumbers = [1, 2, 3, 4, 5, 6, 7],
paddedArray = [],
len = arrayOfNumbers.length;
while(len--) {
paddedArray[len] = ('0000' + arrayOfNumbers[len]).slice(-4);
}
If you don't know how the maximum padding size based on the numbers inside the array.
var arrayOfNumbers = [1, 2, 3, 4, 5, 6, 7, 49095],
paddedArray = [],
len = arrayOfNumbers.length;
// Search the highest number
var arrayMax = Function.prototype.apply.bind(Math.max, null),
// Get that string length
padSize = (arrayMax(arrayOfNumbers) + '').length,
// Create a Padding string
padStr = new Array(padSize).join(0);
// And after you have all this static values cached start the loop.
while(len--) {
paddedArray[len] = (padStr + arrayOfNumbers[len]).slice(-padSize); // substr(-padSize)
}
console.log(paddedArray);
/*
0: "00001"
1: "00002"
2: "00003"
3: "00004"
4: "00005"
5: "00006"
6: "00007"
7: "49095"
*/
padding string has been inplemented in new javascript version.
str.padStart(targetLength [, padString])
https://developer.mozilla.org/es/docs/Web/JavaScript/Referencia/Objetos_globales/String/padStart
If you want your own function check this example:
const myString = 'Welcome to my house';
String.prototype.padLeft = function(times = 0, str = ' ') {
return (Array(times).join(str) + this);
}
console.log(myString.padLeft(12, ':'));
//:::::::::::Welcome to my house
Here is a build in method you can use -
str1.padStart(2, '0')
Here's a simple function that I use.
var pad=function(num,field){
var n = '' + num;
var w = n.length;
var l = field.length;
var pad = w < l ? l-w : 0;
return field.substr(0,pad) + n;
};
For example:
pad (20,' '); // 20
pad (321,' '); // 321
pad (12345,' '); //12345
pad ( 15,'00000'); //00015
pad ( 999,'*****'); //**999
pad ('cat','_____'); //__cat
A short way:
(x=>(new Array(int-x.length+1)).join(char)+x)(String)
Example:
(x=>(new Array(6-x.length+1)).join("0")+x)("1234")
return: "001234"
Here is a simple answer in basically one line of code.
var value = 35 // the numerical value
var x = 5 // the minimum length of the string
var padded = ("00000" + value).substr(-x);
Make sure the number of characters in you padding, zeros here, is at least as many as your intended minimum length. So really, to put it into one line, to get a result of "00035" in this case is:
var padded = ("00000" + 35).substr(-5);
ES7 is just drafts and proposals right now, but if you wanted to track compatibility with the specification, your pad functions need:
Multi-character pad support.
Don't truncate the input string
Pad defaults to space
From my polyfill library, but apply your own due diligence for prototype extensions.
// Tests
'hello'.lpad(4) === 'hello'
'hello'.rpad(4) === 'hello'
'hello'.lpad(10) === ' hello'
'hello'.rpad(10) === 'hello '
'hello'.lpad(10, '1234') === '41234hello'
'hello'.rpad(10, '1234') === 'hello12341'
String.prototype.lpad || (String.prototype.lpad = function(length, pad)
{
if(length < this.length)
return this;
pad = pad || ' ';
let str = this;
while(str.length < length)
{
str = pad + str;
}
return str.substr( -length );
});
String.prototype.rpad || (String.prototype.rpad = function(length, pad)
{
if(length < this.length)
return this;
pad = pad || ' ';
let str = this;
while(str.length < length)
{
str += pad;
}
return str.substr(0, length);
});
Array manipulations are really slow compared to simple string concat. Of course, benchmark for your use case.
function(string, length, pad_char, append) {
string = string.toString();
length = parseInt(length) || 1;
pad_char = pad_char || ' ';
while (string.length < length) {
string = append ? string+pad_char : pad_char+string;
}
return string;
};
A variant of #Daniel LaFavers' answer.
var mask = function (background, foreground) {
bg = (new String(background));
fg = (new String(foreground));
bgl = bg.length;
fgl = fg.length;
bgs = bg.substring(0, Math.max(0, bgl - fgl));
fgs = fg.substring(Math.max(0, fgl - bgl));
return bgs + fgs;
};
For example:
mask('00000', 11 ); // '00011'
mask('00011','00' ); // '00000'
mask( 2 , 3 ); // '3'
mask('0' ,'111'); // '1'
mask('fork' ,'***'); // 'f***'
mask('_____','dog'); // '__dog'
If you don't mind including a utility library, lodash library has _.pad, _.padLeft and _.padRight functions.
I think its better to avoid recursion because its costly.
function padLeft(str,size,padwith) {
if(size <= str.length) {
// not padding is required.
return str;
} else {
// 1- take array of size equal to number of padding char + 1. suppose if string is 55 and we want 00055 it means we have 3 padding char so array size should be 3 + 1 (+1 will explain below)
// 2- now join this array with provided padding char (padwith) or default one ('0'). so it will produce '000'
// 3- now append '000' with orginal string (str = 55), will produce 00055
// why +1 in size of array?
// it is a trick, that we are joining an array of empty element with '0' (in our case)
// if we want to join items with '0' then we should have at least 2 items in the array to get joined (array with single item doesn't need to get joined).
// <item>0<item>0<item>0<item> to get 3 zero we need 4 (3+1) items in array
return Array(size-str.length+1).join(padwith||'0')+str
}
}
alert(padLeft("59",5) + "\n" +
padLeft("659",5) + "\n" +
padLeft("5919",5) + "\n" +
padLeft("59879",5) + "\n" +
padLeft("5437899",5));
It's 2014, and I suggest a JavaScript string-padding function. Ha!
Bare-bones: right-pad with spaces
function pad (str, length) {
var padding = (new Array(Math.max(length - str.length + 1, 0))).join(" ");
return str + padding;
}
Fancy: pad with options
/**
* #param {*} str Input string, or any other type (will be converted to string)
* #param {number} length Desired length to pad the string to
* #param {Object} [opts]
* #param {string} [opts.padWith=" "] Character to use for padding
* #param {boolean} [opts.padLeft=false] Whether to pad on the left
* #param {boolean} [opts.collapseEmpty=false] Whether to return an empty string if the input was empty
* #returns {string}
*/
function pad(str, length, opts) {
var padding = (new Array(Math.max(length - (str + "").length + 1, 0))).join(opts && opts.padWith || " "),
collapse = opts && opts.collapseEmpty && !(str + "").length;
return collapse ? "" : opts && opts.padLeft ? padding + str : str + padding;
}
Usage (fancy):
pad("123", 5);
// Returns "123 "
pad(123, 5);
// Returns "123 " - non-string input
pad("123", 5, { padWith: "0", padLeft: true });
// Returns "00123"
pad("", 5);
// Returns " "
pad("", 5, { collapseEmpty: true });
// Returns ""
pad("1234567", 5);
// Returns "1234567"
/**************************************************************************************************
Pad a string to pad_length fillig it with pad_char.
By default the function performs a left pad, unless pad_right is set to true.
If the value of pad_length is negative, less than, or equal to the length of the input string, no padding takes place.
**************************************************************************************************/
if(!String.prototype.pad)
String.prototype.pad = function(pad_char, pad_length, pad_right)
{
var result = this;
if( (typeof pad_char === 'string') && (pad_char.length === 1) && (pad_length > this.length) )
{
var padding = new Array(pad_length - this.length + 1).join(pad_char); //thanks to http://stackoverflow.com/questions/202605/repeat-string-javascript/2433358#2433358
result = (pad_right ? result + padding : padding + result);
}
return result;
}
And then you can do:
alert( "3".pad("0", 3) ); //shows "003"
alert( "hi".pad(" ", 3) ); //shows " hi"
alert( "hi".pad(" ", 3, true) ); //shows "hi "
If you just want a very simple hacky one-liner to pad, just make a string of the desired padding character of the desired max padding length and then substring it to the length of what you want to pad.
Example: padding the string store in e with spaces to 25 characters long.
var e = "hello"; e = e + " ".substring(e.length)
Result: "hello "
If you want to do the same with a number as input just call .toString() on it before.
A friend asked about using a JavaScript function to pad left. It turned into a little bit of an endeavor between some of us in chat to code golf it. This was the result:
function l(p,t,v){
v+="";return v.length>=t?v:l(p,t,p+v);
}
It ensures that the value to be padded is a string, and then if it isn't the length of the total desired length it will pad it once and then recurse. Here is what it looks like with more logical naming and structure
function padLeft(pad, totalLength, value){
value = value.toString();
if( value.length >= totalLength ){
return value;
}else{
return padLeft(pad, totalLength, pad + value);
}
}
The example we were using was to ensure that numbers were padded with 0 to the left to make a max length of 6. Here is an example set:
function l(p,t,v){v+="";return v.length>=t?v:l(p,t,p+v);}
var vals = [6451,123,466750];
var pad = l(0,6,vals[0]);// pad with 0's, max length 6
var pads = vals.map(function(i){ return l(0,6,i) });
document.write(pads.join("<br />"));
A little late, but thought I might share anyway. I found it useful to add a prototype extension to Object. That way I can pad numbers and strings, left or right. I have a module with similar utilities I include in my scripts.
// include the module in your script, there is no need to export
var jsAddOns = require('<path to module>/jsAddOns');
~~~~~~~~~~~~ jsAddOns.js ~~~~~~~~~~~~
/*
* method prototype for any Object to pad it's toString()
* representation with additional characters to the specified length
*
* #param padToLength required int
* entire length of padded string (original + padding)
* #param padChar optional char
* character to use for padding, default is white space
* #param padLeft optional boolean
* if true padding added to left
* if omitted or false, padding added to right
*
* #return padded string or
* original string if length is >= padToLength
*/
Object.prototype.pad = function(padToLength, padChar, padLeft) {
// get the string value
s = this.toString()
// default padToLength to 0
// if omitted, original string is returned
padToLength = padToLength || 0;
// default padChar to empty space
padChar = padChar || ' ';
// ignore padding if string too long
if (s.length >= padToLength) {
return s;
}
// create the pad of appropriate length
var pad = Array(padToLength - s.length).join(padChar);
// add pad to right or left side
if (padLeft) {
return pad + s;
} else {
return s + pad;
}
};
Never insert data somewhere (especially not at beginning, like str = pad + str;), since the data will be reallocated everytime. Append always at end!
Don't pad your string in the loop. Leave it alone and build your pad string first. In the end concatenate it with your main string.
Don't assign padding string each time (like str += pad;). It is much faster to append the padding string to itself and extract first x-chars (the parser can do this efficiently if you extract from first char). This is exponential growth, which means that it wastes some memory temporarily (you should not do this with extremely huge texts).
if (!String.prototype.lpad) {
String.prototype.lpad = function(pad, len) {
while (pad.length < len) {
pad += pad;
}
return pad.substr(0, len-this.length) + this;
}
}
if (!String.prototype.rpad) {
String.prototype.rpad = function(pad, len) {
while (pad.length < len) {
pad += pad;
}
return this + pad.substr(0, len-this.length);
}
}
Here is a JavaScript function that adds a specified number of paddings with a custom symbol. The function takes three parameters.
padMe --> string or number to left pad
pads --> number of pads
padSymble --> custom symbol, default is "0"
function leftPad(padMe, pads, padSymble) {
if(typeof padMe === "undefined") {
padMe = "";
}
if (typeof pads === "undefined") {
pads = 0;
}
if (typeof padSymble === "undefined") {
padSymble = "0";
}
var symble = "";
var result = [];
for(var i=0; i < pads; i++) {
symble += padSymble;
}
var length = symble.length - padMe.toString().length;
result = symble.substring(0, length);
return result.concat(padMe.toString());
}
Here are some results:
> leftPad(1)
"1"
> leftPad(1, 4)
"0001"
> leftPad(1, 4, "0")
"0001"
> leftPad(1, 4, "#")
"###1"
Yet another take at with combination of a couple of solutions:
/**
* pad string on left
* #param {number} number of digits to pad, default is 2
* #param {string} string to use for padding, default is '0' *
* #returns {string} padded string
*/
String.prototype.paddingLeft = function (b, c) {
if (this.length > (b||2))
return this + '';
return (this || c || 0) + '', b = new Array((++b || 3) - this.length).join(c || 0), b + this
};
/**
* pad string on right
* #param {number} number of digits to pad, default is 2
* #param {string} string to use for padding, default is '0' *
* #returns {string} padded string
*/
String.prototype.paddingRight = function (b, c) {
if (this.length > (b||2))
return this + '';
return (this||c||0) + '', b = new Array((++b || 3) - this.length).join(c || 0), this + b
};
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;
}
I have this script which is working in formatting my currency but now its not fixing my value to 0 decimal places, Im new to javascript so could anybody explain where and hwo I'd do this? Thanks
function FormatNumberBy3(num, decpoint, sep) {
// check for missing parameters and use defaults if so
if (arguments.length == 2) {
sep = ",";
}
if (arguments.length == 1) {
sep = ",";
decpoint = ".";
}
// need a string for operations
num = num.toString();
// separate the whole number and the fraction if possible
a = num.split(decpoint);
x = a[0]; // decimal
y = a[1]; // fraction
z = "";
if (typeof(x) != "undefined") {
// reverse the digits. regexp works from left to right.
for (i=x.length-1;i>=0;i--)
z += x.charAt(i);
// add seperators. but undo the trailing one, if there
z = z.replace(/(\d{3})/g, "$1" + sep);
if (z.slice(-sep.length) == sep)
z = z.slice(0, -sep.length);
x = "";
// reverse again to get back the number
for (i=z.length-1;i>=0;i--)
x += z.charAt(i);
// add the fraction back in, if it was there
if (typeof(y) != "undefined" && y.length > 0)
x += decpoint + y;
}
return x;
}
I may be misunderstanding your question, but it sounds like you want Math.floor()
http://www.javascripter.net/faq/mathfunc.htm
You can round numbers mathematically, which is much, much faster and much, much, much simpler. There are literally hundreds of examples of this algorithm all over this site alone.
function round(number, places)
{
var multiplicator = Math.pow(10, places);
return Math.round(number * multiplicator) / multiplicator;
}
alert(round(123.456, 0)); // alerts "123"
This function lets you round to an arbitrary decimal place. If you want to round to the nearest integer, you may simply use Math.round(x).
Math.round rounds a decimal number to the nearest integer. Assuming you want to keep 3 decimal places, all you have to do is multiply the decimal number by 1000 (which is 10 power 3), round to the nearest integer, then divide by 1000. This is what this snippet does.
Your function appears to work for me... I assume the result of passing "1000" as an argument should be "1,000"? I'm not sure I understand what you mean by "fixing [your] value to 0 decimal places".
Here's a function I wrote to format numbers (like your above function does) for an API once:
function num_format(str) {
if (typeof str !== 'string' || /[^\d,\.e+-]/.test(str)) {
if (typeof str === 'number') {
str = str.toString();
} else {
throw new TypeError("Argument is not a string with a number in it");
}
}
var reg = /(\d+)(\d{3})/;
str = str.split(".");
while (reg.test(str[0])) {
str[0] = str[0].replace(reg, "$1,$2");
}
return str.join(".");
}
Remove the error checking code and it becomes quite a short little function (five lines). Pass it any number and it will format it correctly, although I didn't allow the option of specifying the separators.
As for rounding numbers to arbitrary decimal places, I suggest seeing zneak's answer above. Then would simply do:
num_format(round("100000.12341234", 2));
Which would give a result of "100,000.12".
This function should be capable of all various scenarios:
function formatNumber(num, precision, sep, thousand, addTrailing0) {
if (isNaN(num))
return "";
if (isNaN(precision))
precision = 2;
if (typeof addTrailing0 == "undefined")
addTrailing0 = true;
var factor = Math.pow(10, precision);
num = String(Math.round(num * factor) / factor);
if (addTrailing0 && precision > 0) {
if (num.indexOf(".") < 0)
num += ".";
for (var length = num.substr(num.indexOf(".")+1).length; length < precision; ++length)
num += "0";
}
var parts = num.split(".");
parts[0] = parts[0].split("").reverse().join("").replace(/(\d{3})(?=\d)/g, "$1" + (thousand || ",")).split("").reverse().join("");
num = parts[0] + (parts.length > 1 ? (sep || ".") + parts[1] : "");
//debug:
document.write(num + "<br />");
return num;
}
Examples:
formatNumber(32432342342.3574, 0); // 32,432,342,342
formatNumber(32432342342.3574, 2); // 32,432,342,342.36
formatNumber(1342.525423, 4, ".", ","); // 1,342.5254
formatNumber(1342.525423, 2, ",", "."); // 1.342,53
formatNumber(1342.525423); // 1,342.53
formatNumber(1342.525423, 0); // 1,343
formatNumber(342.525423, 8); // 342.52542300
formatNumber(42.525423, 8, null, null, false); // 42.525423
formatNumber(2.5, 3, ",", "."); // 2,500
Working Live Demo at http://jsfiddle.net/roberkules/xCqqh/
var number = 1310;
should be left alone.
var number = 120;
should be changed to "0120";
var number = 10;
should be changed to "0010";
var number = 7;
should be changed to "0007";
In all modern browsers you can use
numberStr.padStart(4, "0");
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padStart
function zeroPad(num) {
return num.toString().padStart(4, "0");
}
var numbers = [1310, 120, 10, 7];
numbers.forEach(
function(num) {
var paddedNum = zeroPad(num);
console.log(paddedNum);
}
);
function pad_with_zeroes(number, length) {
var my_string = '' + number;
while (my_string.length < length) {
my_string = '0' + my_string;
}
return my_string;
}
try these:
('0000' + number).slice(-4);
or
(number+'').padStart(4,'0');
Here's another way. Comes from something I did that needs to be done thousands of times on a page load. It's pretty CPU efficient to hard code a string of zeroes one time, and chop as many as you need for the pad as many times as needed. I do really like the power of 10 method -- that's pretty flexible.
Anyway, this is as efficient as I could come up with:
For the original question, CHOOSE ONE of the cases...
var number = 1310;
var number = 120;
var number = 10;
var number = 7;
then
// only needs to happen once
var zeroString = "00000";
// one assignment gets the padded number
var paddedNum = zeroString.substring((number + "").length, 4) + bareNum;
//output
alert("The padded number string is: " + paddedNum);
Of course you still need to validate the input. Because this ONLY works reliably under the following conditions:
Number of zeroes in the zeroString is desired_length + 1
Number of digits in your starting number is less than or equal to your desired length
Backstory:
I have a case that needs a fixed length (14 digit) zero-padded number. I wanted to see how basic I could make this. It's run tens of thousands of times on a page load, so efficiency matters. It's not quite re-usable as-is, and it's a bit inelegant. Except that it is very very simple.
For desired n digits padded string, this method requires a string of (at least) n+1 zeroes. Index 0 is the first character in the string, which won't ever be used, so really, it could be anything.
Note also that string.substring() is different from string.substr()!
var bareNum = 42 + '';
var zeroString = "000000000000000";
var paddedNum = zeroString.substring(bareNumber.length, 14) + bareNum
This pulls zeroes from zeroString starting at the position matching the length of the string, and continues to get zeroes to the necessary length of 14. As long as that "14" in the third line is a lower integer than the number of characters in zeroString, it will work.
function pad(n, len) {
return (new Array(len + 1).join('0') + n).slice(-len);
}
might not work in old IE versions.
//to: 0 - to left, 1 - to right
String.prototype.pad = function(_char, len, to) {
if (!this || !_char || this.length >= len) {
return this;
}
to = to || 0;
var ret = this;
var max = (len - this.length)/_char.length + 1;
while (--max) {
ret = (to) ? ret + _char : _char + ret;
}
return ret;
};
Usage:
someString.pad(neededChars, neededLength)
Example:
'332'.pad('0', 6); //'000332'
'332'.pad('0', 6, 1); //'332000'
An approach I like is to add 10^N to the number, where N is the number of zeros you want. Treat the resultant number as a string and slice off the zeroth digit. Of course, you'll want to be careful if your input number might be larger than your pad length, but it's still much faster than the loop method:
// You want to pad four places:
>>> var N = Math.pow(10, 4)
>>> var number = 1310
>>> number < N ? ("" + (N + number)).slice(1) : "" + number
"1310"
>>> var number = 120
>>> number < N ? ("" + (N + number)).slice(1) : "" + number
"0120"
>>> var number = 10
>>> number < N ? ("" + (N + number)).slice(1) : "" + number
"0010"
…
etc. You can make this into a function easily enough:
/**
* Pad a number with leading zeros to "pad" places:
*
* #param number: The number to pad
* #param pad: The maximum number of leading zeros
*/
function padNumber(number, pad) {
var N = Math.pow(10, pad);
return number < N ? ("" + (N + number)).slice(1) : "" + number
}
I wrote a general function for this. It takes an input control and pad length as input.
function padLeft(input, padLength) {
var num = $("#" + input).val();
$("#" + input).val(('0'.repeat(padLength) + num).slice(-padLength));
}
With RegExp/JavaScript:
var number = 7;
number = ('0000'+number).match(/\d{4}$/);
console.log(number);
With Function/RegExp/JavaScript:
var number = 7;
function padFix(n) {
return ('0000'+n).match(/\d{4}$/);
}
console.log(padFix(number));
No loop, no functions
let n = "" + 100;
let x = ("0000000000" + n).substring(n.length);//add your amount of zeros
alert(x + "-" + x.length);
Nate as the best way I found, it's just way too long to read. So I provide you with 3 simples solutions.
1. So here's my simplification of Nate's answer.
//number = 42
"0000".substring(number.toString().length, 4) + number;
2. Here's a solution that make it more reusable by using a function that takes the number and the desired length in parameters.
function pad_with_zeroes(number, len) {
var zeroes = "0".repeat(len);
return zeroes.substring(number.toString().length, len) + number;
}
// Usage: pad_with_zeroes(42,4);
// Returns "0042"
3. Here's a third solution, extending the Number prototype.
Number.prototype.toStringMinLen = function(len) {
var zeroes = "0".repeat(len);
return zeroes.substring(self.toString().length, len) + self;
}
//Usage: tmp=42; tmp.toStringMinLen(4)
Use String.JS librairy function padLeft:
S('123').padLeft(5, '0').s --> 00123
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.