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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)
}

Javascript adding zeros to the beginning of a string (max length 4 chars)

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

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 to convert decimal to hexadecimal in JavaScript

How do you convert decimal values to their hexadecimal equivalent in JavaScript?

Convert a number to a hexadecimal string with:
hexString = yourNumber.toString(16);
And reverse the process with:
yourNumber = parseInt(hexString, 16);

If you need to handle things like bit fields or 32-bit colors, then you need to deal with signed numbers. The JavaScript function toString(16) will return a negative hexadecimal number which is usually not what you want. This function does some crazy addition to make it a positive number.
function decimalToHexString(number)
{
if (number < 0)
{
number = 0xFFFFFFFF + number + 1;
}
return number.toString(16).toUpperCase();
}
console.log(decimalToHexString(27));
console.log(decimalToHexString(48.6));

The code below will convert the decimal value d to hexadecimal. It also allows you to add padding to the hexadecimal result. So 0 will become 00 by default.
function decimalToHex(d, padding) {
var hex = Number(d).toString(16);
padding = typeof (padding) === "undefined" || padding === null ? padding = 2 : padding;
while (hex.length < padding) {
hex = "0" + hex;
}
return hex;
}

function toHex(d) {
return ("0"+(Number(d).toString(16))).slice(-2).toUpperCase()
}

For completeness, if you want the two's-complement hexadecimal representation of a negative number, you can use the zero-fill-right shift >>> operator. For instance:
> (-1).toString(16)
"-1"
> ((-2)>>>0).toString(16)
"fffffffe"
There is however one limitation: JavaScript bitwise operators treat their operands as a sequence of 32 bits, that is, you get the 32-bits two's complement.

With padding:
function dec2hex(i) {
return (i+0x10000).toString(16).substr(-4).toUpperCase();
}

The accepted answer did not take into account single digit returned hexadecimal codes. This is easily adjusted by:
function numHex(s)
{
var a = s.toString(16);
if ((a.length % 2) > 0) {
a = "0" + a;
}
return a;
}
and
function strHex(s)
{
var a = "";
for (var i=0; i<s.length; i++) {
a = a + numHex(s.charCodeAt(i));
}
return a;
}
I believe the above answers have been posted numerous times by others in one form or another. I wrap these in a toHex() function like so:
function toHex(s)
{
var re = new RegExp(/^\s*(\+|-)?((\d+(\.\d+)?)|(\.\d+))\s*$/);
if (re.test(s)) {
return '#' + strHex( s.toString());
}
else {
return 'A' + strHex(s);
}
}
Note that the numeric regular expression came from 10+ Useful JavaScript Regular Expression Functions to improve your web applications efficiency.
Update: After testing this thing several times I found an error (double quotes in the RegExp), so I fixed that. HOWEVER! After quite a bit of testing and having read the post by almaz - I realized I could not get negative numbers to work.
Further - I did some reading up on this and since all JavaScript numbers are stored as 64 bit words no matter what - I tried modifying the numHex code to get the 64 bit word. But it turns out you can not do that. If you put "3.14159265" AS A NUMBER into a variable - all you will be able to get is the "3", because the fractional portion is only accessible by multiplying the number by ten(IE:10.0) repeatedly. Or to put that another way - the hexadecimal value of 0xF causes the floating point value to be translated into an integer before it is ANDed which removes everything behind the period. Rather than taking the value as a whole (i.e.: 3.14159265) and ANDing the floating point value against the 0xF value.
So the best thing to do, in this case, is to convert the 3.14159265 into a string and then just convert the string. Because of the above, it also makes it easy to convert negative numbers because the minus sign just becomes 0x26 on the front of the value.
So what I did was on determining that the variable contains a number - just convert it to a string and convert the string. This means to everyone that on the server side you will need to unhex the incoming string and then to determine the incoming information is numeric. You can do that easily by just adding a "#" to the front of numbers and "A" to the front of a character string coming back. See the toHex() function.
Have fun!
After another year and a lot of thinking, I decided that the "toHex" function (and I also have a "fromHex" function) really needed to be revamped. The whole question was "How can I do this more efficiently?" I decided that a to/from hexadecimal function should not care if something is a fractional part but at the same time it should ensure that fractional parts are included in the string.
So then the question became, "How do you know you are working with a hexadecimal string?". The answer is simple. Use the standard pre-string information that is already recognized around the world.
In other words - use "0x". So now my toHex function looks to see if that is already there and if it is - it just returns the string that was sent to it. Otherwise, it converts the string, number, whatever. Here is the revised toHex function:
/////////////////////////////////////////////////////////////////////////////
// toHex(). Convert an ASCII string to hexadecimal.
/////////////////////////////////////////////////////////////////////////////
toHex(s)
{
if (s.substr(0,2).toLowerCase() == "0x") {
return s;
}
var l = "0123456789ABCDEF";
var o = "";
if (typeof s != "string") {
s = s.toString();
}
for (var i=0; i<s.length; i++) {
var c = s.charCodeAt(i);
o = o + l.substr((c>>4),1) + l.substr((c & 0x0f),1);
}
return "0x" + o;
}
This is a very fast function that takes into account single digits, floating point numbers, and even checks to see if the person is sending a hex value over to be hexed again. It only uses four function calls and only two of those are in the loop. To un-hex the values you use:
/////////////////////////////////////////////////////////////////////////////
// fromHex(). Convert a hex string to ASCII text.
/////////////////////////////////////////////////////////////////////////////
fromHex(s)
{
var start = 0;
var o = "";
if (s.substr(0,2).toLowerCase() == "0x") {
start = 2;
}
if (typeof s != "string") {
s = s.toString();
}
for (var i=start; i<s.length; i+=2) {
var c = s.substr(i, 2);
o = o + String.fromCharCode(parseInt(c, 16));
}
return o;
}
Like the toHex() function, the fromHex() function first looks for the "0x" and then it translates the incoming information into a string if it isn't already a string. I don't know how it wouldn't be a string - but just in case - I check. The function then goes through, grabbing two characters and translating those in to ASCII characters. If you want it to translate Unicode, you will need to change the loop to going by four(4) characters at a time. But then you also need to ensure that the string is NOT divisible by four. If it is - then it is a standard hexadecimal string. (Remember the string has "0x" on the front of it.)
A simple test script to show that -3.14159265, when converted to a string, is still -3.14159265.
<?php
echo <<<EOD
<html>
<head><title>Test</title>
<script>
var a = -3.14159265;
alert( "A = " + a );
var b = a.toString();
alert( "B = " + b );
</script>
</head>
<body>
</body>
</html>
EOD;
?>
Because of how JavaScript works in respect to the toString() function, all of those problems can be eliminated which before were causing problems. Now all strings and numbers can be converted easily. Further, such things as objects will cause an error to be generated by JavaScript itself. I believe this is about as good as it gets. The only improvement left is for W3C to just include a toHex() and fromHex() function in JavaScript.

Without the loop:
function decimalToHex(d) {
var hex = Number(d).toString(16);
hex = "000000".substr(0, 6 - hex.length) + hex;
return hex;
}
// Or "#000000".substr(0, 7 - hex.length) + hex;
// Or whatever
// *Thanks to MSDN
Also isn't it better not to use loop tests that have to be evaluated?
For example, instead of:
for (var i = 0; i < hex.length; i++){}
have
for (var i = 0, var j = hex.length; i < j; i++){}

Combining some of these good ideas for an RGB-value-to-hexadecimal function (add the # elsewhere for HTML/CSS):
function rgb2hex(r,g,b) {
if (g !== undefined)
return Number(0x1000000 + r*0x10000 + g*0x100 + b).toString(16).substring(1);
else
return Number(0x1000000 + r[0]*0x10000 + r[1]*0x100 + r[2]).toString(16).substring(1);
}

Constrained/padded to a set number of characters:
function decimalToHex(decimal, chars) {
return (decimal + Math.pow(16, chars)).toString(16).slice(-chars).toUpperCase();
}

For anyone interested, here's a JSFiddle comparing most of the answers given to this question.
And here's the method I ended up going with:
function decToHex(dec) {
return (dec + Math.pow(16, 6)).toString(16).substr(-6)
}
Also, bear in mind that if you're looking to convert from decimal to hex for use in CSS as a color data type, you might instead prefer to extract the RGB values from the decimal and use rgb().
For example (JSFiddle):
let c = 4210330 // your color in decimal format
let rgb = [(c & 0xff0000) >> 16, (c & 0x00ff00) >> 8, (c & 0x0000ff)]
// Vanilla JS:
document.getElementById('some-element').style.color = 'rgb(' + rgb + ')'
// jQuery:
$('#some-element').css('color', 'rgb(' + rgb + ')')
This sets #some-element's CSS color property to rgb(64, 62, 154).

var number = 3200;
var hexString = number.toString(16);
The 16 is the radix and there are 16 values in a hexadecimal number :-)

function dec2hex(i)
{
var result = "0000";
if (i >= 0 && i <= 15) { result = "000" + i.toString(16); }
else if (i >= 16 && i <= 255) { result = "00" + i.toString(16); }
else if (i >= 256 && i <= 4095) { result = "0" + i.toString(16); }
else if (i >= 4096 && i <= 65535) { result = i.toString(16); }
return result
}

If you want to convert a number to a hexadecimal representation of an RGBA color value, I've found this to be the most useful combination of several tips from here:
function toHexString(n) {
if(n < 0) {
n = 0xFFFFFFFF + n + 1;
}
return "0x" + ("00000000" + n.toString(16).toUpperCase()).substr(-8);
}

AFAIK comment 57807 is wrong and should be something like:
var hex = Number(d).toString(16);
instead of
var hex = parseInt(d, 16);
function decimalToHex(d, padding) {
var hex = Number(d).toString(16);
padding = typeof (padding) === "undefined" || padding === null ? padding = 2 : padding;
while (hex.length < padding) {
hex = "0" + hex;
}
return hex;
}

And if the number is negative?
Here is my version.
function hexdec (hex_string) {
hex_string=((hex_string.charAt(1)!='X' && hex_string.charAt(1)!='x')?hex_string='0X'+hex_string : hex_string);
hex_string=(hex_string.charAt(2)<8 ? hex_string =hex_string-0x00000000 : hex_string=hex_string-0xFFFFFFFF-1);
return parseInt(hex_string, 10);
}

As the accepted answer states, the easiest way to convert from decimal to hexadecimal is var hex = dec.toString(16). However, you may prefer to add a string conversion, as it ensures that string representations like "12".toString(16) work correctly.
// Avoids a hard-to-track-down bug by returning `c` instead of `12`
(+"12").toString(16);
To reverse the process you may also use the solution below, as it is even shorter.
var dec = +("0x" + hex);
It seems to be slower in Google Chrome and Firefox, but is significantly faster in Opera. See http://jsperf.com/hex-to-dec.

I'm doing conversion to hex string in a pretty large loop, so I tried several techniques in order to find the fastest one. My requirements were to have a fixed-length string as a result, and encode negative values properly (-1 => ff..f).
Simple .toString(16) didn't work for me since I needed negative values to be properly encoded. The following code is the quickest I've tested so far on 1-2 byte values (note that symbols defines the number of output symbols you want to get, that is for 4-byte integer it should be equal to 8):
var hex = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'b', 'c', 'd', 'e', 'f'];
function getHexRepresentation(num, symbols) {
var result = '';
while (symbols--) {
result = hex[num & 0xF] + result;
num >>= 4;
}
return result;
}
It performs faster than .toString(16) on 1-2 byte numbers and slower on larger numbers (when symbols >= 6), but still should outperform methods that encode negative values properly.

Converting hex color numbers to hex color strings:
A simple solution with toString and ES6 padStart for converting hex color numbers to hex color strings.
const string = `#${color.toString(16).padStart(6, '0')}`;
For example:
0x000000 will become #000000
0xFFFFFF will become #FFFFFF
Check this example in a fiddle here

How to convert decimal to hexadecimal in JavaScript
I wasn't able to find a brutally clean/simple decimal to hexadecimal conversion that didn't involve a mess of functions and arrays ... so I had to make this for myself.
function DecToHex(decimal) { // Data (decimal)
length = -1; // Base string length
string = ''; // Source 'string'
characters = [ '0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F' ]; // character array
do { // Grab each nibble in reverse order because JavaScript has no unsigned left shift
string += characters[decimal & 0xF]; // Mask byte, get that character
++length; // Increment to length of string
} while (decimal >>>= 4); // For next character shift right 4 bits, or break on 0
decimal += 'x'; // Convert that 0 into a hex prefix string -> '0x'
do
decimal += string[length];
while (length--); // Flip string forwards, with the prefixed '0x'
return (decimal); // return (hexadecimal);
}
/* Original: */
D = 3678; // Data (decimal)
C = 0xF; // Check
A = D; // Accumulate
B = -1; // Base string length
S = ''; // Source 'string'
H = '0x'; // Destination 'string'
do {
++B;
A& = C;
switch(A) {
case 0xA: A='A'
break;
case 0xB: A='B'
break;
case 0xC: A='C'
break;
case 0xD: A='D'
break;
case 0xE: A='E'
break;
case 0xF: A='F'
break;
A = (A);
}
S += A;
D >>>= 0x04;
A = D;
} while(D)
do
H += S[B];
while (B--)
S = B = A = C = D; // Zero out variables
alert(H); // H: holds hexadecimal equivalent

You can do something like this in ECMAScript 6:
const toHex = num => (num).toString(16).toUpperCase();

If you are looking for converting Large integers i.e. Numbers greater than Number.MAX_SAFE_INTEGER -- 9007199254740991, then you can use the following code
const hugeNumber = "9007199254740991873839" // Make sure its in String
const hexOfHugeNumber = BigInt(hugeNumber).toString(16);
console.log(hexOfHugeNumber)

To sum it all up;
function toHex(i, pad) {
if (typeof(pad) === 'undefined' || pad === null) {
pad = 2;
}
var strToParse = i.toString(16);
while (strToParse.length < pad) {
strToParse = "0" + strToParse;
}
var finalVal = parseInt(strToParse, 16);
if ( finalVal < 0 ) {
finalVal = 0xFFFFFFFF + finalVal + 1;
}
return finalVal;
}
However, if you don't need to convert it back to an integer at the end (i.e. for colors), then just making sure the values aren't negative should suffice.

I haven't found a clear answer, without checks if it is negative or positive, that uses two's complement (negative numbers included). For that, I show my solution to one byte:
((0xFF + number +1) & 0x0FF).toString(16);
You can use this instruction to any number bytes, only you add FF in respective places. For example, to two bytes:
((0xFFFF + number +1) & 0x0FFFF).toString(16);
If you want cast an array integer to string hexadecimal:
s = "";
for(var i = 0; i < arrayNumber.length; ++i) {
s += ((0xFF + arrayNumber[i] +1) & 0x0FF).toString(16);
}

In case you're looking to convert to a 'full' JavaScript or CSS representation, you can use something like:
numToHex = function(num) {
var r=((0xff0000&num)>>16).toString(16),
g=((0x00ff00&num)>>8).toString(16),
b=(0x0000ff&num).toString(16);
if (r.length==1) { r = '0'+r; }
if (g.length==1) { g = '0'+g; }
if (b.length==1) { b = '0'+b; }
return '0x'+r+g+b; // ('#' instead of'0x' for CSS)
};
var dec = 5974678;
console.log( numToHex(dec) ); // 0x5b2a96

This is based on Prestaul and Tod's solutions. However, this is a generalisation that accounts for varying size of a variable (e.g. Parsing signed value from a microcontroller serial log).
function decimalToPaddedHexString(number, bitsize)
{
let byteCount = Math.ceil(bitsize/8);
let maxBinValue = Math.pow(2, bitsize)-1;
/* In node.js this function fails for bitsize above 32bits */
if (bitsize > 32)
throw "number above maximum value";
/* Conversion to unsigned form based on */
if (number < 0)
number = maxBinValue + number + 1;
return "0x"+(number >>> 0).toString(16).toUpperCase().padStart(byteCount*2, '0');
}
Test script:
for (let n = 0 ; n < 64 ; n++ ) {
let s=decimalToPaddedHexString(-1, n);
console.log(`decimalToPaddedHexString(-1,${(n+"").padStart(2)}) = ${s.padStart(10)} = ${("0b"+parseInt(s).toString(2)).padStart(34)}`);
}
Test results:
decimalToPaddedHexString(-1, 0) = 0x0 = 0b0
decimalToPaddedHexString(-1, 1) = 0x01 = 0b1
decimalToPaddedHexString(-1, 2) = 0x03 = 0b11
decimalToPaddedHexString(-1, 3) = 0x07 = 0b111
decimalToPaddedHexString(-1, 4) = 0x0F = 0b1111
decimalToPaddedHexString(-1, 5) = 0x1F = 0b11111
decimalToPaddedHexString(-1, 6) = 0x3F = 0b111111
decimalToPaddedHexString(-1, 7) = 0x7F = 0b1111111
decimalToPaddedHexString(-1, 8) = 0xFF = 0b11111111
decimalToPaddedHexString(-1, 9) = 0x01FF = 0b111111111
decimalToPaddedHexString(-1,10) = 0x03FF = 0b1111111111
decimalToPaddedHexString(-1,11) = 0x07FF = 0b11111111111
decimalToPaddedHexString(-1,12) = 0x0FFF = 0b111111111111
decimalToPaddedHexString(-1,13) = 0x1FFF = 0b1111111111111
decimalToPaddedHexString(-1,14) = 0x3FFF = 0b11111111111111
decimalToPaddedHexString(-1,15) = 0x7FFF = 0b111111111111111
decimalToPaddedHexString(-1,16) = 0xFFFF = 0b1111111111111111
decimalToPaddedHexString(-1,17) = 0x01FFFF = 0b11111111111111111
decimalToPaddedHexString(-1,18) = 0x03FFFF = 0b111111111111111111
decimalToPaddedHexString(-1,19) = 0x07FFFF = 0b1111111111111111111
decimalToPaddedHexString(-1,20) = 0x0FFFFF = 0b11111111111111111111
decimalToPaddedHexString(-1,21) = 0x1FFFFF = 0b111111111111111111111
decimalToPaddedHexString(-1,22) = 0x3FFFFF = 0b1111111111111111111111
decimalToPaddedHexString(-1,23) = 0x7FFFFF = 0b11111111111111111111111
decimalToPaddedHexString(-1,24) = 0xFFFFFF = 0b111111111111111111111111
decimalToPaddedHexString(-1,25) = 0x01FFFFFF = 0b1111111111111111111111111
decimalToPaddedHexString(-1,26) = 0x03FFFFFF = 0b11111111111111111111111111
decimalToPaddedHexString(-1,27) = 0x07FFFFFF = 0b111111111111111111111111111
decimalToPaddedHexString(-1,28) = 0x0FFFFFFF = 0b1111111111111111111111111111
decimalToPaddedHexString(-1,29) = 0x1FFFFFFF = 0b11111111111111111111111111111
decimalToPaddedHexString(-1,30) = 0x3FFFFFFF = 0b111111111111111111111111111111
decimalToPaddedHexString(-1,31) = 0x7FFFFFFF = 0b1111111111111111111111111111111
decimalToPaddedHexString(-1,32) = 0xFFFFFFFF = 0b11111111111111111111111111111111
Thrown: 'number above maximum value'
Note: Not too sure why it fails above 32 bitsize

rgb(255, 255, 255) // returns FFFFFF
rgb(255, 255, 300) // returns FFFFFF
rgb(0,0,0) // returns 000000
rgb(148, 0, 211) // returns 9400D3
function rgb(...values){
return values.reduce((acc, cur) => {
let val = cur >= 255 ? 'ff' : cur <= 0 ? '00' : Number(cur).toString(16);
return acc + (val.length === 1 ? '0'+val : val);
}, '').toUpperCase();
}

Arbitrary precision
This solution take on input decimal string, and return hex string. A decimal fractions are supported. Algorithm
split number to sign (s), integer part (i) and fractional part (f) e.g for -123.75 we have s=true, i=123, f=75
integer part to hex:
if i='0' stop
get modulo: m=i%16 (in arbitrary precision)
convert m to hex digit and put to result string
for next step calc integer part i=i/16 (in arbitrary precision)
fractional part
count fractional digits n
multiply k=f*16 (in arbitrary precision)
split k to right part with n digits and put them to f, and left part with rest of digits and put them to d
convert d to hex and add to result.
finish when number of result fractional digits is enough
// #param decStr - string with non-negative integer
// #param divisor - positive integer
function dec2HexArbitrary(decStr, fracDigits=0) {
// Helper: divide arbitrary precision number by js number
// #param decStr - string with non-negative integer
// #param divisor - positive integer
function arbDivision(decStr, divisor)
{
// algorithm https://www.geeksforgeeks.org/divide-large-number-represented-string/
let ans='';
let idx = 0;
let temp = +decStr[idx];
while (temp < divisor) temp = temp * 10 + +decStr[++idx];
while (decStr.length > idx) {
ans += (temp / divisor)|0 ;
temp = (temp % divisor) * 10 + +decStr[++idx];
}
if (ans.length == 0) return "0";
return ans;
}
// Helper: calc module of arbitrary precision number
// #param decStr - string with non-negative integer
// #param mod - positive integer
function arbMod(decStr, mod) {
// algorithm https://www.geeksforgeeks.org/how-to-compute-mod-of-a-big-number/
let res = 0;
for (let i = 0; i < decStr.length; i++)
res = (res * 10 + +decStr[i]) % mod;
return res;
}
// Helper: multiply arbitrary precision integer by js number
// #param decStr - string with non-negative integer
// #param mult - positive integer
function arbMultiply(decStr, mult) {
let r='';
let m=0;
for (let i = decStr.length-1; i >=0 ; i--) {
let n = m+mult*(+decStr[i]);
r= (i ? n%10 : n) + r
m= n/10|0;
}
return r;
}
// dec2hex algorithm starts here
let h= '0123456789abcdef'; // hex 'alphabet'
let m= decStr.match(/-?(.*?)\.(.*)?/) || decStr.match(/-?(.*)/); // separate sign,integer,ractional
let i= m[1].replace(/^0+/,'').replace(/^$/,'0'); // integer part (without sign and leading zeros)
let f= (m[2]||'0').replace(/0+$/,'').replace(/^$/,'0'); // fractional part (without last zeros)
let s= decStr[0]=='-'; // sign
let r=''; // result
if(i=='0') r='0';
while(i!='0') { // integer part
r=h[arbMod(i,16)]+r;
i=arbDivision(i,16);
}
if(fracDigits) r+=".";
let n = f.length;
for(let j=0; j<fracDigits; j++) { // frac part
let k= arbMultiply(f,16);
f = k.slice(-n);
let d= k.slice(0,k.length-n);
r+= d.length ? h[+d] : '0';
}
return (s?'-':'')+r;
}
// -----------
// TESTS
// -----------
let tests = [
["0",2],
["000",2],
["123",0],
["-123",0],
["00.000",2],
["255.75",5],
["-255.75",5],
["127.999",32],
];
console.log('Input Standard Abitrary');
tests.forEach(t=> {
let nonArb = (+t[0]).toString(16).padEnd(17,' ');
let arb = dec2HexArbitrary(t[0],t[1]);
console.log(t[0].padEnd(10,' '), nonArb, arb);
});
// Long Example (40 digits after dot)
let example = "123456789012345678901234567890.09876543210987654321"
console.log(`\nLong Example:`);
console.log('dec:',example);
console.log('hex: ',dec2HexArbitrary(example,40));

The problem basically how many padding zeros to expect.
If you expect string 01 and 11 from Number 1 and 17. it's better to use Buffer as a bridge, with which number is turn into bytes, and then the hex is just an output format of it. And the bytes organization is well controlled by Buffer functions, like writeUInt32BE, writeInt16LE, etc.
import { Buffer } from 'buffer';
function toHex(n) { // 4byte
const buff = Buffer.alloc(4);
buff.writeInt32BE(n);
return buff.toString('hex');
}
> toHex(1)
'00000001'
> toHex(17)
'00000011'
> toHex(-1)
'ffffffff'
> toHex(-1212)
'fffffb44'
> toHex(1212)
'000004bc'

Here's my solution:
hex = function(number) {
return '0x' + Math.abs(number).toString(16);
}
The question says: "How to convert decimal to hexadecimal in JavaScript". While, the question does not specify that the hexadecimal string should begin with a 0x prefix, anybody who writes code should know that 0x is added to hexadecimal codes to distinguish hexadecimal codes from programmatic identifiers and other numbers (1234 could be hexadecimal, decimal, or even octal).
Therefore, to correctly answer this question, for the purpose of script-writing, you must add the 0x prefix.
The Math.abs(N) function converts negatives to positives, and as a bonus, it doesn't look like somebody ran it through a wood-chipper.
The answer I wanted, would have had a field-width specifier, so we could for example show 8/16/32/64-bit values the way you would see them listed in a hexadecimal editing application. That, is the actual, correct answer.