[javascript] How can I pad a value with leading zeros?

What is the recommended way to zerofill a value in JavaScript? I imagine I could build a custom function to pad zeros on to a typecasted value, but I'm wondering if there is a more direct way to do this?

Note: By "zerofilled" I mean it in the database sense of the word (where a 6-digit zerofilled representation of the number 5 would be "000005").

I have 2 solutions. The first is the real basic one to bad a number with zeros if you know that's all you want to do quickly.

The second will also pad negative numbers - same as the "clever" solution that seems to have the best answer score.

Here goes:

// One liner simple mode!_x000D_
// (always ensure the number is 6 digits long knowing its never negative)_x000D_
var mynum = 12;_x000D_
var mynum2 = "0".repeat((n=6-mynum.toString().length)>0?n:0)+mynum;_x000D_
alert("One liner to pad number to be 6 digits long = Was "+mynum+" Now "+mynum2);_x000D_
_x000D_
// As a function which will also pad negative numbers_x000D_
// Yes, i could do a check to only take "-" into account_x000D_
// if it was passed in as an integer and not a string but_x000D_
// for this, i haven't._x000D_
// _x000D_
// @s The string or integer to pad_x000D_
// @l The minumum length of @s_x000D_
// @c The character to pad with_x000D_
// @return updated string_x000D_
function padme(s,l,c){_x000D_
    s = s.toString();_x000D_
    c = c.toString();_x000D_
    m = s.substr(0,1)=="-";_x000D_
    return (m?"-":"")+c.repeat((n=l-(s.length-(m?1:0)))>0?n:0)+s.substr((m?1:0));_x000D_
}_x000D_
alert("pad -12 to ensure it is 8 digits long = "+padme(-12,8,0));_x000D_
alert("pad 'hello' with 'x' to ensure it is 12 digits long = "+padme('hello',12,'x'));

I came up with an absurd one-liner while writing a numeric base converter. Didn't see anything quite like it in the other answers, so here goes:

// This is cursed
function p(i,w,z){z=z||0;w=w||8;i+='';var o=i.length%w;return o?[...Array(w-o).fill(z),...i].join(''):i;}

console.log(p(8675309));        // Default: pad w/ 0 to 8 digits
console.log(p(525600, 10));     // Pad to 10 digits
console.log(p(69420, 10, 'X')); // Pad w/ X to 10 digits
console.log(p(8675309, 4));     // Pad to next 4 digits
console.log(p(12345678));       // Don't pad if you ain't gotta pad

Or, in a form that doesn't quite as readily betray that I've sold my soul to the Black Perl:

function pad(input, width, zero) {
    zero = zero || 0; width = width || 8;  // Defaults
    input += '';                           // Convert input to string first
    
    var overflow = input.length % width    // Do we overflow?
    if (overflow) {                        // Yep!  Let's pad it...
        var needed = width - overflow;     // ...to the next boundary...
        var zeroes = Array(needed);        // ...with an array...
        zeroes = zeroes.fill(zero);        // ...full of our zero character...
        var output = [...zeroes,...input]; // ...and concat those zeroes to input...
        output = output.join('');          // ...and finally stringify.
    } else {
        var output = input;                // We don't overflow; no action needed :)
    }
    
    return output;                         // Done!
}

One thing that sets this apart from the other answers is that it takes a modulo of the number's length to the target width rather than a simple greater-than check. This is handy if you want to make sure the resulting length is some multiple of a target width (e.g. you need the output to be either 5 or 10 characters long).

No idea how well it performs, but hey, at least it's already minified!

My contribution:

I'm assuming you want the total string length to include the 'dot'. If not it's still simple to rewrite to add an extra zero if the number is a float.

padZeros = function (num, zeros) {
        return (((num < 0) ? "-" : "") + Array(++zeros - String(Math.abs(num)).length).join("0") + Math.abs(num));
    }

Yet another version :

function zPad(s,n){
    return (new Array(n+1).join('0')+s).substr(-Math.max(n,s.toString().length));
}

First parameter is any real number, second parameter is a positive integer specifying the minimum number of digits to the left of the decimal point and third parameter is an optional positive integer specifying the number if digits to the right of the decimal point.

function zPad(n, l, r){
    return(a=String(n).match(/(^-?)(\d*)\.?(\d*)/))?a[1]+(Array(l).join(0)+a[2]).slice(-Math.max(l,a[2].length))+('undefined'!==typeof r?(0<r?'.':'')+(a[3]+Array(r+1).join(0)).slice(0,r):a[3]?'.'+a[3]:''):0
}

so

           zPad(6, 2) === '06'
          zPad(-6, 2) === '-06'
       zPad(600.2, 2) === '600.2'
        zPad(-600, 2) === '-600'
         zPad(6.2, 3) === '006.2'
        zPad(-6.2, 3) === '-006.2'
      zPad(6.2, 3, 0) === '006'
        zPad(6, 2, 3) === '06.000'
    zPad(600.2, 2, 3) === '600.200'
zPad(-600.1499, 2, 3) === '-600.149'

The power of Math!

x = integer to pad
y = number of zeroes to pad

function zeroPad(x, y)
{
   y = Math.max(y-1,0);
   var n = (x / Math.pow(10,y)).toFixed(y);
   return n.replace('.','');  
}

In all modern browsers you can use

numberStr.padStart(numberLength, "0");

function zeroFill(num, numLength) {_x000D_
  var numberStr = num.toString();_x000D_
_x000D_
  return numberStr.padStart(numLength, "0");_x000D_
}_x000D_
_x000D_
var numbers = [0, 1, 12, 123, 1234, 12345];_x000D_
_x000D_
numbers.forEach(_x000D_
  function(num) {_x000D_
    var numString = num.toString();_x000D_
    _x000D_
    var paddedNum = zeroFill(numString, 5);_x000D_
_x000D_
    console.log(paddedNum);_x000D_
  }_x000D_
);

Here is the MDN reference https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padStart

function zFill(n,l){
    return 
      (l > n.toString().length) ? 
        ( (Array(l).join('0') + n).slice(-l) ) : n;
}

A simple short recursive function to achieve your proposal:

function padleft (YourNumber, OutputLength){
    if (YourNumber.length >= OutputLength) {
        return YourNumber;
    } else {
        return padleft("0" +YourNumber, OutputLength);
    }
}

• YourNumber is the input number.
• OutputLength is the preferred output number length (with 0 padding left).

This function will add 0 on the left if your input number length is shorter than the wanted output number length.

Even later to the party.

function zfill(num, len) {
  return(0 > num ? "-" : "") + (Math.pow(10, len) <= Math.abs(num) ? "0" + Math.abs(num) : Math.pow(10, len) + Math.abs(num)).toString().substr(1)
}

This handles negatives and situations where the number is longer than the field width. And floating-point.

With ES6+ JavaScript:

You can "zerofill a number" with something like the following function:

/**
 * @param number The number
 * @param minLength Minimal length for your string with leading zeroes
 * @return Your formatted string
 */
function zerofill(nb, minLength) {
    // Convert your number to string.
    let nb2Str = nb.toString()

    // Guess the number of zeroes you will have to write.
    let nbZeroes = Math.max(0, minLength - nb2Str.length)

    // Compute your result.
    return `${ '0'.repeat(nbZeroes) }${ nb2Str }`
}

console.log(zerofill(5, 6))    // Displays "000005"

With ES2017+:

/**
 * @param number The number
 * @param minLength Minimal length for your string with leading zeroes
 * @return Your formatted string
 */
const zerofill = (nb, minLength) => nb.toString().padStart(minLength, '0')

console.log(zerofill(5, 6))    // Displays "000005"

If the fill number is known in advance not to exceed a certain value, there's another way to do this with no loops:

var fillZeroes = "00000000000000000000";  // max number of zero fill ever asked for in global

function zeroFill(number, width) {
    // make sure it's a string
    var input = number + "";  
    var prefix = "";
    if (input.charAt(0) === '-') {
        prefix = "-";
        input = input.slice(1);
        --width;
    }
    var fillAmt = Math.max(width - input.length, 0);
    return prefix + fillZeroes.slice(0, fillAmt) + input;
}

Test cases here: http://jsfiddle.net/jfriend00/N87mZ/

Here's a quick function I came up with to do the job. If anyone has a simpler approach, feel free to share!

function zerofill(number, length) {
    // Setup
    var result = number.toString();
    var pad = length - result.length;

    while(pad > 0) {
        result = '0' + result;
        pad--;
    }

    return result;
}

Unfortunately, there are a lot of needless complicated suggestions for this problem, typically involving writing your own function to do math or string manipulation or calling a third-party utility. However, there is a standard way of doing this in the base JavaScript library with just one line of code. It might be worth wrapping this one line of code in a function to avoid having to specify parameters that you never want to change like the local name or style.

var amount = 5;

var text = amount.toLocaleString('en-US',
{
    style: 'decimal',
    minimumIntegerDigits: 3,
    useGrouping: false
});

This will produce the value of "005" for text. You can also use the toLocaleString function of Number to pad zeros to the right side of the decimal point.

var amount = 5;

var text = amount.toLocaleString('en-US',
{
    style: 'decimal',
    minimumFractionDigits: 2,
    useGrouping: false
});

This will produce the value of "5.00" for text. Change useGrouping to true to use comma separators for thousands.

Note that using toLocaleString() with locales and options arguments is standardized separately in ECMA-402, not in ECMAScript. As of today, some browsers only implement basic support, i.e. toLocaleString() may ignore any arguments.

Complete Example

I actually had to come up with something like this recently. I figured there had to be a way to do it without using loops.

This is what I came up with.

function zeroPad(num, numZeros) {
    var n = Math.abs(num);
    var zeros = Math.max(0, numZeros - Math.floor(n).toString().length );
    var zeroString = Math.pow(10,zeros).toString().substr(1);
    if( num < 0 ) {
        zeroString = '-' + zeroString;
    }

    return zeroString+n;
}

Then just use it providing a number to zero pad:

> zeroPad(50,4);
"0050"

If the number is larger than the padding, the number will expand beyond the padding:

> zeroPad(51234, 3);
"51234"

Decimals are fine too!

> zeroPad(51.1234, 4);
"0051.1234"

If you don't mind polluting the global namespace you can add it to Number directly:

Number.prototype.leftZeroPad = function(numZeros) {
    var n = Math.abs(this);
    var zeros = Math.max(0, numZeros - Math.floor(n).toString().length );
    var zeroString = Math.pow(10,zeros).toString().substr(1);
    if( this < 0 ) {
        zeroString = '-' + zeroString;
    }

    return zeroString+n;
}

And if you'd rather have decimals take up space in the padding:

Number.prototype.leftZeroPad = function(numZeros) {
    var n = Math.abs(this);
    var zeros = Math.max(0, numZeros - n.toString().length );
    var zeroString = Math.pow(10,zeros).toString().substr(1);
    if( this < 0 ) {
        zeroString = '-' + zeroString;
    }

    return zeroString+n;
}

Cheers!

XDR came up with a logarithmic variation that seems to perform better.

WARNING: This function fails if num equals zero (e.g. zeropad(0, 2))

function zeroPad (num, numZeros) {
    var an = Math.abs (num);
    var digitCount = 1 + Math.floor (Math.log (an) / Math.LN10);
    if (digitCount >= numZeros) {
        return num;
    }
    var zeroString = Math.pow (10, numZeros - digitCount).toString ().substr (1);
    return num < 0 ? '-' + zeroString + an : zeroString + an;
}

Speaking of performance, tomsmeding compared the top 3 answers (4 with the log variation). Guess which one majorly outperformed the other two? :)

Maybe I am to naive, but I think that this works in one simple and efficient line of code (for positive numbers):

padded = (value+Math.pow(10,total_length)+"").slice(1)

As long as you keep your length OK according to you set of values (as in any zero padding), this should work.

The steps are:

1. Add the power of 10 with the correct number of 0's [69+1000 = 1069]
2. Convert to string with +"" [1069 => "1069"]
3. Slice the first 1, which resulted of first multiplication ["1069" => "069"]

For natural listings (files, dirs...) is quite useful.

My solution

Number.prototype.PadLeft = function (length, digit) {
    var str = '' + this;
    while (str.length < length) {
        str = (digit || '0') + str;
    }
    return str;
};

Usage

var a = 567.25;
a.PadLeft(10); // 0000567.25

var b = 567.25;
b.PadLeft(20, '2'); // 22222222222222567.25

function uint_zerofill(num, width) {
    var pad = ''; num += '';
    for (var i = num.length; i < width; i++)
        pad += '0';
    return pad + num;
}

My little contribution with this topic (https://gist.github.com/lucasferreira/a881606894dde5568029):

/* Autor: Lucas Ferreira - http://blog.lucasferreira.com | Usage: fz(9) or fz(100, 7) */
function fz(o, s) {
    for(var s=Math.max((+s||2),(n=""+Math.abs(o)).length); n.length<s; (n="0"+n));
    return (+o < 0 ? "-" : "") + n;
};

Usage:

fz(9) & fz(9, 2) == "09"
fz(-3, 2) == "-03"
fz(101, 7) == "0000101"

I know, it's a pretty dirty function, but it's fast and works even with negative numbers ;)

ECMAScript 2017: use padStart or padEnd

'abc'.padStart(10);         // "       abc"
'abc'.padStart(10, "foo");  // "foofoofabc"
'abc'.padStart(6,"123465"); // "123abc"

More info:

• https://github.com/tc39/proposal-string-pad-start-end
• https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padStart

A simple elegant solution, where n is the number and l is the length.

function nFill (n, l) {return (l>n.toString().length)?((Array(l).join('0')+n).slice(-l)):n;}

This keeps the length if it is over desired, as not to alter the number.

n = 500;_x000D_
_x000D_
console.log(nFill(n, 5));_x000D_
console.log(nFill(n, 2));_x000D_
_x000D_
function nFill (n, l) {return (l>n.toString().length)?((Array(l).join('0')+n).slice(-l)):n;}

Why not use recursion?

function padZero(s, n) {
    s = s.toString(); // in case someone passes a number
    return s.length >= n ? s : padZero('0' + s, n);
}

This is an angular provider that I wrote, which makes use of @profitehlolz 's answer but employs memoization so that commonly used pad length-pad character combinations will not invoke the array build join needlessly:

angular.module('stringUtilities', [])
    .service('stringFunctions', [function() {
        this.padMemo={ };
        this.padLeft=function(inputString,padSize,padCharacter) {

            var memoKey=padSize+""+padCharacter;

            if(!this.padMemo[memoKey]) {

                this.padMemo[memoKey]= new Array(1 + padSize).join(padCharacter);
            }

           var pad=this.padMemo[memoKey];
           return (pad + inputString).slice(-pad.length);
       };
}]);

Some monkeypatching also works

String.prototype.padLeft = function (n, c) {
  if (isNaN(n))
    return null;
  c = c || "0";
  return (new Array(n).join(c).substring(0, this.length-n)) + this; 
};
var paddedValue = "123".padLeft(6); // returns "000123"
var otherPadded = "TEXT".padLeft(8, " "); // returns "    TEXT"

just wanted to make the comment (but i don't have enough points) that the highest voted answer fails with negative numbers and decimals

function padNumber(n,pad) {
    p = Math.pow(10,pad);
    a = Math.abs(n);
    g = (n<0);
    return (a < p) ?  ((g ? '-' : '') + (p+a).toString().substring(1)) : n;
}

padNumber( -31.235, 5);

"-00031.235"

This is the ES6 solution.

function pad(num, len) {_x000D_
  return '0'.repeat(len - num.toString().length) + num;_x000D_
}_x000D_
alert(pad(1234,6));

I didn't see anyone point out the fact that when you use String.prototype.substr() with a negative number it counts from the right.

A one liner solution to the OP's question, a 6-digit zerofilled representation of the number 5, is:

console.log(("00000000" + 5).substr(-6));

Generalizing we'll get:

function pad(num, len) { return ("00000000" + num).substr(-len) };_x000D_
_x000D_
console.log(pad(5, 6));_x000D_
console.log(pad(45, 6));_x000D_
console.log(pad(345, 6));_x000D_
console.log(pad(2345, 6));_x000D_
console.log(pad(12345, 6));

variable-length padding function

function addPaddingZeroes(value, nLength)
{
    var sValue = value + ''; //converts to string

    if(sValue.length>=nLength)
        return sValue;
    else
    {
        for(var nZero = 0; nZero < nLength; nZero++)
            sValue = "0" + sValue;
        return (sValue).substring(nLength - sValue.length, nLength);    
    }
}

Just for fun, here's my version of a pad function:

function pad(num, len) {
  return Array(len + 1 - num.toString().length).join('0') + num;
}

It also won't truncate numbers longer than the padding length

Here's a little trick I think is cool:

(2/10000).toString().split(".")[1]
"0002"
(52/10000).toString().split(".")[1]
"0052"

Not that this question needs more answers, but I thought I would add the simple lodash version of this.

_.padLeft(number, 6, '0')

I am using this simple approach

var input = 1000; //input any number
var len = input.toString().length;
for (i = 1; i < input; i++) {
  console.log("MyNumber_" + ('000000000000000' + i).slice(-len));
}

This one is less native, but may be the fastest...

zeroPad = function (num, count) {
    var pad = (num + '').length - count;
    while(--pad > -1) {
        num = '0' + num;
    }
    return num;
};

function numPadding (padding,i) {
    return padding.substr(0, padding.length - (Math.floor(i).toString().length)) + Math.floor(i );
}

numPadding("000000000",234); -> "000000234"

or

function numPadding (number, paddingChar,i) {
    var padding = new Array(number + 1).join(paddingChar);
    return padding.substr(0, padding.length - (Math.floor(i).toString().length)) + Math.floor(i );
}

numPadding(8 ,"0", 234); -> "00000234";

Here's what I used to pad a number up to 7 characters.

("0000000" + number).slice(-7)

This approach will probably suffice for most people.

Edit: If you want to make it more generic you can do this:

("0".repeat(padding) + number).slice(-padding)

Edit 2: Note that since ES2017 you can use String.prototype.padStart:

number.toString().padStart(padding, "0")

A simple one for my use case (to fill milliseconds never > 999) You can adjust the number of zeros for yours or use a more generic way if required.

/**
 * @val integer
 * @zeros padding
 */
function zeroFill(val, zeros)
{
    var str = val.toString();
    if (str.length >= zeros)
        return str;
    str = "000" + str;
    return str.substring(str.length - zeros);
}

Just an another solution, but I think it's more legible.

function zeroFill(text, size)_x000D_
{_x000D_
  while (text.length < size){_x000D_
   text = "0" + text;_x000D_
  }_x000D_
  _x000D_
  return text;_x000D_
}

Ok, ok, I can't believe I'm bothering to submit another answer to this one, but I think my approach is a little different. The reason I needed to pad a number was to display it in a <pre> element (part of an on-screen log), so it's ultimately going to be a string anyway. Instead of doing any math, I wrote a simple function to overlay a string value on a mask string:

function overlayr(m, s) {
  return m.length > s.length ? m.substr(0, m.length - s.length) + s : s;
}

The benefit of this is that I can use it for all sorts of string alignment tasks. To call it, just pass in the mask and number as a string:

> overlayr('00000', (5).toString())
< "00005"

As an added bonus, it deals with overflows correctly:

> overlayr('00000', (555555).toString())
< "555555"

And of course it's not limited to 0 padding:

> overlayr('*****', (55).toString())
< "***55"

A little math can give you a one-line function:

function zeroFill( number, width ) {
  return Array(width - parseInt(Math.log(number)/Math.LN10) ).join('0') + number;
}

That's assuming that number is an integer no wider than width. If the calling routine can't make that guarantee, the function will need to make some checks:

function zeroFill( number, width ) {
    var n = width - parseInt(Math.log(number)/Math.LN10);
    return (n < 0) ? '' + number : Array(n).join('0') + number;
}

The following provides a quick and fast solution:

function numberPadLeft(num , max, padder = "0"){_x000D_
     return "" == (num += "") ? "" :_x000D_
     ( dif = max - num.length, dif > 0 ?_x000D_
     padder.repeat(dif < 0 ? 0 : dif) + num :_x000D_
     num )_x000D_
}

I use this snipet to get a 5 digits representation

(value+100000).toString().slice(-5) // "00123" with value=123

was here looking for a standard. had the same idea as Paul and Jonathan... theirs are super cute, here's a horrible-cute version:

function zeroPad(n,l,i){
    return (i=n/Math.pow(10,l))*i>1?''+n:i.toFixed(l).replace('0.','');
}

works too (we're assuming integers, yes?)...

> zeroPad(Math.pow(2, 53), 20);
'00009007199254740992'
> zeroPad(-Math.pow(2, 53), 20);
'-00009007199254740992'
> zeroPad(Math.pow(2, 53), 10);
'9007199254740992'
> zeroPad(-Math.pow(2, 53), 10);
'-9007199254740992'

Our tests were bogus because mine had a typo.

zeroPad = function (num, count) {
    return ((num / Math.pow(10, count)) + '').substr(2);
};

Paul's is the fastest, but I think .substr is faster than .slice even if it is one character more ;)

exports.pad = (num, length) => "0".repeat(length - num.toString().length) + num;

To pad at the end of the number, use num.toFixed

for example:

  document.getElementById('el').value = amt.toFixed(2);

It's the simplest solution i've found, and it works.

Simple way. You could add string multiplication for the pad and turn it into a function.

var pad = "000000";
var n = '5';
var result = (pad+n).slice(-pad.length);

As a function,

function paddy(num, padlen, padchar) {
    var pad_char = typeof padchar !== 'undefined' ? padchar : '0';
    var pad = new Array(1 + padlen).join(pad_char);
    return (pad + num).slice(-pad.length);
}
var fu = paddy(14, 5); // 00014
var bar = paddy(2, 4, '#'); // ###2

Mnah... I have not seen a "ultimate" answer to this issue and if you are facing the same challenge I must save you some time by saying that saddly there's not built-in function for that on JavaScript; but there's this awesome function in PHP that does a great job on padding strings as well as numbers with single character or arbitrary strings so after some time of banging my head for not having the right tool on JS [mostly for zerofillin' numbers and usually for trimming strings to fit a fixed length] and excessive coding work I decided to write my own function that does the same ["almost the same", read on for detail] that the dream PHP function but in comfortable client-side JavaScript.

function str_pad(input,pad_length,pad_string,pad_type){
    var input=input.toString();
    var output="";
    if((input.length>pad_length)&&(pad_type=='STR_PAD_RIGHT')){var output=input.slice(0,pad_length);}
    else if((input.length>pad_length)&&(pad_type=='STR_PAD_LEFT')){var output=input.slice(input.length-pad_length,input.length);}
    else if((input.length<pad_length)&&(pad_type=='STR_PAD_RIGHT')){
        var caracteresNecesarios=pad_length-input.length;
        var rellenoEnteros=Math.floor(caracteresNecesarios/pad_string.length);
        var rellenoParte=caracteresNecesarios%pad_string.length;
        var output=input;
        for(var i=0;i<rellenoEnteros;i++){var output=output+pad_string;};
        var output=output+pad_string.slice(0,rellenoParte);
    }
    else if((input.length<pad_length)&&(pad_type=='STR_PAD_LEFT')){
        var caracteresNecesarios=pad_length-input.length;
        var rellenoEnteros=Math.floor(caracteresNecesarios/pad_string.length);
        var rellenoParte=caracteresNecesarios%pad_string.length;
        var output="";
        for(var i=0;i<rellenoEnteros;i++){var output=output+pad_string;};
        var output=output+pad_string.slice(0,rellenoParte);
        var output=output+input;
    }
    else if(input.length==pad_length){var output=input;};
    return output;
};

The only thing that my function does not do is the STR_PAD_BOTH behavior that I could add with some time and a more comfortable keyboard. You might call the function and test it; bet you'll love it if you don't mind that inner code uses one or two words in Spanish... not big deal I think. I did not added comments for "watermarking" my coding so you can seamless use it in your work nor I compressed the code for enhanced readability. Use it and test it like this and spread the code:

alert("str_pad('murcielago',20,'123','STR_PAD_RIGHT')="+str_pad('murcielago',20,'123','STR_PAD_RIGHT')+'.');

function pad(toPad, padChar, length){
    return (String(toPad).length < length)
        ? new Array(length - String(toPad).length + 1).join(padChar) + String(toPad)
        : toPad;
}

pad(5, 0, 6) = 000005

pad('10', 0, 2) = 10 // don't pad if not necessary

pad('S', 'O', 2) = SO

...etc.

Cheers

I just stumbled upon this post looking for a native solution. Since there isn't a built-in solution, here's my take on it:

function zerofill(number, width) {
    var num = '';
    while (width-- > 0) {
        num += '0';
    }

    return num.slice(0, - (number + '').length) + number + '';
}

The quick and dirty way:

y = (new Array(count + 1 - x.toString().length)).join('0') + x;

For x = 5 and count = 6 you'll have y = "000005"

function numberPadding(n, p) {
  n = n.toString();
  var len = p - n.length;
  if (len > 0) {
    for (var i=0; i < len; i++) {
      n = '0' + n;
    }
  }
  return n;
}

Here a little array solution within a two line function. It checks also if the leading zeros are less than the length of the number string.

function pad(num, z) {
    if (z < (num = num + '').length) return num;
    return Array(++z - num.length).join('0') + num;
}

This method isn't faster, but it's fairly native.

zeroPad = function (num, count) {
    return [Math.pow(10, count - num.toString().length), num].join('').substr(1);
};

If performance is really critical (looping over millions of records), an array of padding strings can be pre-generated, avoiding to do it for each call.

Time complexity: O(1).
Space complexity: O(1).

const zeroPads = Array.from({ length: 10 }, (_, v) => '0'.repeat(v))_x000D_
_x000D_
function zeroPad(num, len) {_x000D_
  const numStr = String(num)_x000D_
  return (zeroPads[len - numStr.length] + numStr)_x000D_
}

I can't believe all the complex answers on here... Just use this:

var zerofilled = ('0000'+n).slice(-4);

I really don't know why, but no one did it in the most obvious way. Here it's my implementation.

Function:

/** Pad a number with 0 on the left */
function zeroPad(number, digits) {
    var num = number+"";
    while(num.length < digits){
        num='0'+num;
    }
    return num;
}

Prototype:

Number.prototype.zeroPad=function(digits){
    var num=this+"";
    while(num.length < digits){
        num='0'+num;
    }
    return(num);
};

Very straightforward, I can't see any way how this can be any simpler. For some reason I've seem many times here on SO, people just try to avoid 'for' and 'while' loops at any cost. Using regex will probably cost way more cycles for such a trivial 8 digit padding.

sprintf.js is a complete open source JavaScript sprintf implementation for the browser and node.js.

Its prototype is simple:

string sprintf(string format , [mixed arg1 [, mixed arg2 [ ,...]]])

I'd like to recommend sprintf module from Alexandru Mara?teanu throughout the solution would simply looks like:

var sprintf = require('sprintf');
var zeroFilled = sprintf('%06d', 5);

console.log(zeroFilled); // 000005

Note: I'm answering this question 6 years later but it seems that this question becomes a "javascript zero leading" reference considering it's high number of views and answers.

ES6 makes this fairly trivial:

function pad (num, length, countSign = true) {
  num = num.toString()
  let negative = num.startsWith('-')
  let numLength = negative && !countSign ? num.length - 1 : num.length
  if (numLength >= length) {
    return num
  } else if (negative) {
    return '-' + '0'.repeat(length - numLength) + num.substr(1)
  } else {
    return '0'.repeat(length - numLength) + num
  }
}

pad(42, 4)          === '0042'
pad(12345, 4)       === '12345'
pad(-123, 4)        === '-100'
pad(-123, 4, false) === '-0100'

I didn't see any answer in this form so here my shot with regex and string manipulation

(Works also for negative and decimal numbers)

Code:

function fillZeroes(n = 0, m = 1) {
  const p = Math.max(1, m);
  return String(n).replace(/\d+/, x => '0'.repeat(Math.max(p - x.length, 0)) + x);
}

Some outputs:

console.log(fillZeroes(6, 2))          // >> '06'
console.log(fillZeroes(1.35, 2))       // >> '01.35'
console.log(fillZeroes(-16, 3))        // >> '-016'
console.log(fillZeroes(-1.456, 3))     // >> '-001.456'
console.log(fillZeroes(-456.53453, 6)) // >> '-000456.53453'
console.log(fillZeroes('Agent 7', 3))  // >> 'Agent 007'

Late to the party here, but I often use this construct for doing ad-hoc padding of some value n, known to be a positive, decimal:

(offset + n + '').substr(1);

Where offset is 10^^digits.

E.g. Padding to 5 digits, where n = 123:

(1e5 + 123 + '').substr(1); // => 00123

The hexidecimal version of this is slightly more verbose:

(0x100000 + 0x123).toString(16).substr(1); // => 00123

Note 1: I like @profitehlolz's solution as well, which is the string version of this, using slice()'s nifty negative-index feature.

The latest way to do this is much simpler:

var number = 2
number.toLocaleString(undefined, {minimumIntegerDigits:2})

output: "02"

A simple function to do it:

function padStr(number, numDigits){
  return 
    (number < 0 ? '-':'') 
    + ((new Array(numDigits + 1).join("0"))
    + Math.abs(number)).slice(-numDigits);
}

function zeroFill(number, width) {
    width -= (number.toString().length - /\./.test(number));
    if (width > 0) {
        return new Array(width + 1).join('0') + number;
    }
    return number + ""; // always return a string
}

Slight changes made to Peter's code. With his code if the input is (1.2, 3) the value returned should be 01.2 but it is returning 1.2. The changes here should correct that.

Don't reinvent the wheel, use underscore string:

jsFiddle

var numToPad = '5';

alert(_.str.pad(numToPad, 6, '0')); // yields: '000005'

If you use Lodash.

var n = 1;_x000D_
_x000D_
alert( _.padLeft(n, 2, 0) ); // 01_x000D_
_x000D_
n = 10;_x000D_
_x000D_
alert( _.padLeft(n, 2, 0) ); // 10

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/3.10.0/lodash.min.js"></script>

If npm is available in your environment some of ready-made packages can be used: www.npmjs.com/browse/keyword/zeropad.

I like zero-fill.

Installation

$ npm install zero-fill

Usage

var zeroFill = require('zero-fill')

zeroFill(4, 1)      // '0001' 
zeroFill(4, 1, '#') // '###1' custom padding
zeroFill(4)(1)      // '0001' partials

The simplest, most straight-forward solution you will find.

function zerofill(number,length) {
    var output = number.toString();
    while(output.length < length) {
      output = '0' + output;
    }
    return output;
}

function zeroPad(num,digits){ return ((num/Math.pow(10,digits))+'').slice(2) } 

Just an FYI, clearer, more readable syntax IMHO

"use strict";
String.prototype.pad = function( len, c, left ) {
    var s = '',
        c = ( c || ' ' ),
        len = Math.max( len, 0 ) - this.length,
        left = ( left || false );
    while( s.length < len ) { s += c };
    return ( left ? ( s + this ) : ( this + s ) );
}
Number.prototype.pad = function( len, c, left ) {
    return String( this ).pad( len, c, left );
}
Number.prototype.lZpad = function( len ) {
    return this.pad( len, '0', true );
}

This also results in less visual and readability glitches of the results than some of the other solutions, which enforce '0' as a character; answering my questions what do I do if I want to pad other characters, or on the other direction (right padding), whilst remaining easy to type, and clear to read. Pretty sure it's also the DRY'est example, with the least code for the actual leading-zero-padding function body (as the other dependent functions are largely irrelevant to the question).

The code is available for comment via gist from this github user (original source of the code) https://gist.github.com/Lewiscowles1986/86ed44f428a376eaa67f

A note on console & script testing, numeric literals seem to need parenthesis, or a variable in order to call methods, so 2.pad(...) will cause an error, whilst (2).pad(0,'#') will not. This is the same for all numbers it seems

i wrote somethin in ecmaScript6 (TypeScript) and perhaps someone can use it:

class Helper {
    /**
     * adds leading 0 and returns string if value is not minSize long, 
     * else returns value as string
     *
     * @param {string|number} value
     * @param {number} minSize
     * @returns {string}
     */
    public static leadingNullString(value: string|number, minSize: number): string {
        if (typeof value == "number") {
            value = "" + value;
        }
        let outString: string = '';
        let counter: number = minSize - value.length;
        if (counter > 0) {
            for (let i = 0; i < counter; i++) {
                outString += '0';
            }
        }
        return (outString + value);
    }
}

Helper.leadingNullString(123, 2); returns "123"

Helper.leadingNullString(5, 2); returns "05"

Helper.leadingNullString(40,2); returns "40"

The ecmaScript4 (JavaScript) transpilation looks like that:

var Helper = (function () {
    function Helper() {
    }
    Helper.leadingNullString = function (value, minSize) {
        if (typeof value == "number") {
            value = "" + value;
        }
        var outString = '';
        var counter = minSize - value.length;
        if (counter > 0) {
            for (var i = 0; i < counter; i++) {
                outString += '0';
            }
        }
        return (outString + value);
    };
    return Helper;
}());

Modern browsers now support padStart, you can simply now do:

string.padStart(maxLength, "0");

Example:

string = "14";
maxLength = 5; // maxLength is the max string length, not max # of fills
res = string.padStart(maxLength, "0");
console.log(res); // prints "00014"

number = 14;
maxLength = 5; // maxLength is the max string length, not max # of fills
res = number.toString().padStart(maxLength, "0");
console.log(res); // prints "00014"

I used

Utilities.formatString("%04d", iThe_TWO_to_FOUR_DIGIT) 

which gives up to 4 leading 0s

NOTE: THIS REQUIRES Google's apps-script Utilities:

https://developers.google.com/apps-script/reference/utilities/utilities#formatstringtemplate-args

After a, long, long time of testing 15 different functions/methods found in this questions answers, I now know which is the best (the most versatile and quickest).

I took 15 functions/methods from the answers to this question and made a script to measure the time taken to execute 100 pads. Each pad would pad the number 9 with 2000 zeros. This may seem excessive, and it is, but it gives you a good idea about the scaling of the functions.

The code I used can be found here: https://gist.github.com/NextToNothing/6325915

Feel free to modify and test the code yourself.

In order to get the most versatile method, you have to use a loop. This is because with very large numbers others are likely to fail, whereas, this will succeed.

So, which loop to use? Well, that would be a while loop. A for loop is still fast, but a while loop is just slightly quicker(a couple of ms) - and cleaner.

Answers like those by Wilco, Aleksandar Toplek or Vitim.us will do the job perfectly.

Personally, I tried a different approach. I tried to use a recursive function to pad the string/number. It worked out better than methods joining an array but, still, didn't work as quick as a for loop.

My function is:

function pad(str, max, padder) {
  padder = typeof padder === "undefined" ? "0" : padder;
  return str.toString().length < max ? pad(padder.toString() + str, max, padder) : str;
}

You can use my function with, or without, setting the padding variable. So like this:

pad(1, 3); // Returns '001'
// - Or -
pad(1, 3, "x"); // Returns 'xx1'

Personally, after my tests, I would use a method with a while loop, like Aleksandar Toplek or Vitim.us. However, I would modify it slightly so that you are able to set the padding string.

So, I would use this code:

function padLeft(str, len, pad) {
    pad = typeof pad === "undefined" ? "0" : pad + "";
    str = str + "";
    while(str.length < len) {
        str = pad + str;
    }
    return str;
}

// Usage
padLeft(1, 3); // Returns '001'
// - Or -
padLeft(1, 3, "x"); // Returns 'xx1'

You could also use it as a prototype function, by using this code:

Number.prototype.padLeft = function(len, pad) {
    pad = typeof pad === "undefined" ? "0" : pad + "";
    var str = this + "";
    while(str.length < len) {
        str = pad + str;
    }
    return str;
}

// Usage
var num = 1;

num.padLeft(3); // Returns '001'
// - Or -
num.padLeft(3, "x"); // Returns 'xx1'