[javascript] Is there a JavaScript function that can pad a string to get to a determined length?

I am in need of a JavaScript function which can take a value and pad it to a given length (I need spaces, but anything would do). I found this:

Code:

String.prototype.pad = function(l, s, t){
    return s || (s = " "), (l -= this.length) > 0 ? (s = new Array(Math.ceil(l / s.length)
        + 1).join(s)).substr(0, t = !t ? l : t == 1 ? 0 : Math.ceil(l / 2))
        + this + s.substr(0, l - t) : this;
};

Example:

var s = "Jonas";
document.write(
    '<h2>S = '.bold(), s, "</h2>",
    'S.pad(20, "[]", 0) = '.bold(), s.pad(20, "[]", 0), "<br />",
    'S.pad(20, "[====]", 1) = '.bold(), s.pad(20, "[====]", 1), "<br />",
    'S.pad(20, "~", 2) = '.bold(), s.pad(20, "~", 2)
);

But I have no idea what the heck it is doing and it doesn't seem to work for me.

/**************************************************************************************************
Pad a string to pad_length fillig it with pad_char.
By default the function performs a left pad, unless pad_right is set to true.

If the value of pad_length is negative, less than, or equal to the length of the input string, no padding takes place.
**************************************************************************************************/
if(!String.prototype.pad)
String.prototype.pad = function(pad_char, pad_length, pad_right) 
{
   var result = this;
   if( (typeof pad_char === 'string') && (pad_char.length === 1) && (pad_length > this.length) )
   {
      var padding = new Array(pad_length - this.length + 1).join(pad_char); //thanks to http://stackoverflow.com/questions/202605/repeat-string-javascript/2433358#2433358
      result = (pad_right ? result + padding : padding + result);
   }
   return result;
}

And then you can do:

alert( "3".pad("0", 3) ); //shows "003"
alert( "hi".pad(" ", 3) ); //shows " hi"
alert( "hi".pad(" ", 3, true) ); //shows "hi "

A short way:

(x=>(new Array(int-x.length+1)).join(char)+x)(String)

Example:

(x=>(new Array(6-x.length+1)).join("0")+x)("1234")

return: "001234"

Using the ECMAScript 6 method String#repeat, a pad function is as simple as:

String.prototype.padLeft = function(char, length) { 
    return char.repeat(Math.max(0, length - this.length)) + this;
}

String#repeat is currently supported in Firefox and Chrome only. for other implementation, one might consider the following simple polyfill:

String.prototype.repeat = String.prototype.repeat || function(n){ 
    return n<=1 ? this : (this + this.repeat(n-1)); 
}

It's 2014, and I suggest a Javascript string-padding function. Ha!

Bare-bones: right-pad with spaces

function pad ( str, length ) {
    var padding = ( new Array( Math.max( length - str.length + 1, 0 ) ) ).join( " " );
    return str + padding;
}

Fancy: pad with options

/**
 * @param {*}       str                         input string, or any other type (will be converted to string)
 * @param {number}  length                      desired length to pad the string to
 * @param {Object}  [opts]
 * @param {string}  [opts.padWith=" "]          char to use for padding
 * @param {boolean} [opts.padLeft=false]        whether to pad on the left
 * @param {boolean} [opts.collapseEmpty=false]  whether to return an empty string if the input was empty
 * @returns {string}
 */
function pad ( str, length, opts ) {
    var padding = ( new Array( Math.max( length - ( str + "" ).length + 1, 0 ) ) ).join( opts && opts.padWith || " " ),
        collapse = opts && opts.collapseEmpty && !( str + "" ).length;
    return collapse ? "" : opts && opts.padLeft ? padding + str : str + padding;
}

Usage (fancy):

pad( "123", 5 );
// returns "123  "

pad( 123, 5 );
// returns "123  " - non-string input

pad( "123", 5, { padWith: "0", padLeft: true } );
// returns "00123"

pad( "", 5 );
// returns "     "

pad( "", 5, { collapseEmpty: true } );
// returns ""

pad( "1234567", 5 );
// returns "1234567"

With ES8, there are two options for padding.

You can check them in the documentation.

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padEnd

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padStart

yet another take at with combination of a couple solutions

/**
 * pad string on left
 * @param {number} number of digits to pad, default is 2
 * @param {string} string to use for padding, default is '0' *
 * @returns {string} padded string
 */
String.prototype.paddingLeft = function (b,c) {
    if (this.length > (b||2))
        return this+'';
  return (this||c||0)+'',b=new Array((++b||3)-this.length).join(c||0),b+this
};

/**
 * pad string on right
 * @param {number} number of digits to pad, default is 2
 * @param {string} string to use for padding, default is '0' *
 * @returns {string} padded string
 */
String.prototype.paddingRight = function (b,c) {
  if (this.length > (b||2))
        return this+'';
  return (this||c||0)+'',b=new Array((++b||3)-this.length).join(c||0),this+b
};    

String.prototype.padLeft = function(pad) {
        var s = Array.apply(null, Array(pad)).map(function() { return "0"; }).join('') + this;
        return s.slice(-1 * Math.max(this.length, pad));
    };

usage:

1. "123".padLeft(2) returns: "123"
2. "12".padLeft(2) returns: "12"
3. "1".padLeft(2) returns: "01"

Try this:-

function leftPad(number) {
     return (number < 9)?'0'+number:number;
}

//call like this
var month=3;
month=leftPad(month);//output:- month=04

Based on the best answers of this question I have made a prototype for String called padLeft (exactly like we have in C#):

String.prototype.padLeft = function (paddingChar, totalWidth) {
    if (this.toString().length >= totalWidth)
        return this.toString();

    var array = new Array(totalWidth); 

    for (i = 0; i < array.length; i++)
        array[i] = paddingChar;

    return (array.join("") + this.toString()).slice(-array.length);
}

Usage:

var str = "12345";
console.log(str.padLeft("0", 10)); //Result is: "0000012345"

JsFiddle

I think its better to avoid recursion because its costly.

function padLeft(str,size,padwith) {_x000D_
 if(size <= str.length) {_x000D_
        // not padding is required._x000D_
  return str;_x000D_
 } else {_x000D_
        // 1- take array of size equal to number of padding char + 1. suppose if string is 55 and we want 00055 it means we have 3 padding char so array size should be 3 + 1 (+1 will explain below)_x000D_
        // 2- now join this array with provided padding char (padwith) or default one ('0'). so it will produce '000'_x000D_
        // 3- now append '000' with orginal string (str = 55), will produce 00055_x000D_
_x000D_
        // why +1 in size of array? _x000D_
        // it is a trick, that we are joining an array of empty element with '0' (in our case)_x000D_
        // if we want to join items with '0' then we should have at least 2 items in the array to get joined (array with single item doesn't need to get joined)._x000D_
        // <item>0<item>0<item>0<item> to get 3 zero we need 4 (3+1) items in array   _x000D_
  return Array(size-str.length+1).join(padwith||'0')+str_x000D_
 }_x000D_
}_x000D_
_x000D_
alert(padLeft("59",5) + "\n" +_x000D_
     padLeft("659",5) + "\n" +_x000D_
     padLeft("5919",5) + "\n" +_x000D_
     padLeft("59879",5) + "\n" +_x000D_
     padLeft("5437899",5));

Here's a simple function that I use.

var pad=function(num,field){
    var n = '' + num;
    var w = n.length;
    var l = field.length;
    var pad = w < l ? l-w : 0;
    return field.substr(0,pad) + n;
};

For example:

pad    (20,'     ');    //   20
pad   (321,'     ');    //  321
pad (12345,'     ');    //12345
pad (   15,'00000');    //00015
pad (  999,'*****');    //**999
pad ('cat','_____');    //__cat  

The key trick in both those solutions is to create an array instance with a given size (one more than the desired length), and then to immediately call the join() method to make a string. The join() method is passed the padding string (spaces probably). Since the array is empty, the empty cells will be rendered as empty strings during the process of joining the array into one result string, and only the padding will remain. It's a really nice technique.

For something like this, I might create a one-line function at the point where it is needed:

var padleft = (s,c,len) => { while(s.length < len) s = c + s; return s; }

Example:

> console.log( padleft( '110', '0', 8) );
> 00000110

1. function
var _padLeft = function(paddingString, width, replacementChar) {
    return paddingString.length >= width ? paddingString : _padLeft(replacementChar + paddingString, width, replacementChar || ' ');
};

2. String prototype
String.prototype.padLeft = function(width, replacementChar) {
        return this.length >= width ? this.toString() : (replacementChar + this).padLeft(width, replacementChar || ' ');
};

3. slice
('00000' + paddingString).slice(-5)

I like to do this in case you ever need to pad with multiple characters or tags (e.g.  ) for display:

$.padStringLeft = function(s, pad, len) {
    if(typeof s !== 'undefined') {
        var c=s.length; while(len > c) {s=pad+s;c++;}
    }
    return s;
}    

$.padStringRight = function(s, pad, len) {
    if(typeof s !== 'undefined') {
        var c=s.length; while(len > c) {s += pad;c++;}
    }
    return s;
}

A faster method

If you are doing this repeatedly, for example to pad values in an array, and performance is a factor, the following approach can give you nearly a 100x advantage in speed (jsPerf) over other solution that are currently discussed on the inter webs. The basic idea is that you are providing the pad function with a fully padded empty string to use as a buffer. The pad function just appends to string to be added to this pre-padded string (one string concat) and then slices or trims the result to the desired length.

function pad(pad, str, padLeft) {
  if (typeof str === 'undefined') 
    return pad;
  if (padLeft) {
    return (pad + str).slice(-pad.length);
  } else {
    return (str + pad).substring(0, pad.length);
  }
}

For example, to zero pad a number to a length of 10 digits,

pad('0000000000',123,true);

To pad a string with whitespace, so the entire string is 255 characters,

var padding = Array(256).join(' '), // make a string of 255 spaces
pad(padding,123,true);

Performance Test

See the jsPerf test here.

And this is faster than ES6 string.repeat by 2x as well, as shown by the revised JsPerf here

Please note that jsPerf is no longer online

Please note that the jsPerf site that we originally used to benchmark the various methods is no longer online. Unfortunately, this means we can't get to those test results. Sad but true.

1. Never insert data somewhere (especially not at beginning, like str = pad + str;), since the data will be reallocated everytime. Append always at end!
2. Don't pad your string in the loop. Leave it alone and build your pad string first. In the end concatenate it with your main string.
3. Don't assign padding string each time (like str += pad;). It is much faster to append the padding string to itself and extract first x-chars (the parser can do this efficiently if you extract from first char). This is exponential growth, which means that it wastes some memory temporarily (you should not do this with extremely huge texts).

if (!String.prototype.lpad) {_x000D_
    String.prototype.lpad = function(pad, len) {_x000D_
        while (pad.length < len) {_x000D_
            pad += pad;_x000D_
        }_x000D_
        return pad.substr(0, len-this.length) + this;_x000D_
    }_x000D_
}_x000D_
_x000D_
if (!String.prototype.rpad) {_x000D_
    String.prototype.rpad = function(pad, len) {_x000D_
        while (pad.length < len) {_x000D_
            pad += pad;_x000D_
        }_x000D_
        return this + pad.substr(0, len-this.length);_x000D_
    }_x000D_
}

Here's a recursive approach to it.

function pad(width, string, padding) { 
  return (width <= string.length) ? string : pad(width, padding + string, padding)
}

An example...

pad(5, 'hi', '0')
=> "000hi"

Array manipulations are really slow compared to simple string concat. Of course, benchmark for your use case.

function(string, length, pad_char, append) {
    string = string.toString();
    length = parseInt(length) || 1;
    pad_char = pad_char || ' ';

    while (string.length < length) {
        string = append ? string+pad_char : pad_char+string;
    }
    return string;
};

If you just want a very simple hacky one-liner to pad, just make a string of the desired padding character of the desired max padding length and then substring it to the length of what you want to pad.

Example: padding the string store in e with spaces to 25 characters long.

var e = "hello"; e = e + "                         ".substring(e.length)

Result: "hello "

If you want to do the same with a number as input just call .toString() on it before.

padding string has been inplemented in new javascript version.

str.padStart(targetLength [, padString])

https://developer.mozilla.org/es/docs/Web/JavaScript/Referencia/Objetos_globales/String/padStart

If you want your own function check this example:

const myString = 'Welcome to my house';
String.prototype.padLeft = function(times = 0, str = ' ') {
    return (Array(times).join(str) + this);
}
console.log(myString.padLeft(12, ':'));
//:::::::::::Welcome to my house

String.prototype.padStart() and String.prototype.padEnd() are currently TC39 candidate proposals: see github.com/tc39/proposal-string-pad-start-end (only available in Firefox as of April 2016; a polyfill is available).

my combination of aboves solutions added to my own, always evolving version :)

//in preperation for ES6
String.prototype.lpad || (String.prototype.lpad = function( length, charOptional )
{
    if (length <= this.length) return this;
    return ( new Array((length||0)+1).join(String(charOptional)||' ') + (this||'') ).slice( -(length||0) );
});


'abc'.lpad(5,'.') == '..abc'
String(5679).lpad(10,0) == '0000005679'
String().lpad(4,'-') == '----' // repeat string

Here's my take

I'm not so sure about it's performance, but I find it much more readable than other options I saw around here...

var replicate = function(len, char) {
  return Array(len+1).join(char || ' ');
};

var padr = function(text, len, char) {
  if (text.length >= len) return text;
  return text + replicate(len-text.length, char);
};

this is my version of function:

function str_pad(str, size, char, right) {
    var s = str + "";
    while (s.length < size) {
        if (right) {
            s = s + char;
        } else {
            s = char + s;
        }
    }
    return s;
}

A little late, but thought I might share anyway. I found it useful to add a prototype extension to Object. That way I can pad numbers and strings, left or right. I have a module with similar utilities I include in my scripts.

// include the module in your script, there is no need to export
var jsAddOns = require('<path to module>/jsAddOns');

~~~~~~~~~~~~ jsAddOns.js ~~~~~~~~~~~~

/* 
 * method prototype for any Object to pad it's toString()
 * representation with additional characters to the specified length
 *
 * @param padToLength required int
 *     entire length of padded string (original + padding)
 * @param padChar optional char
 *     character to use for padding, default is white space
 * @param padLeft optional boolean
 *     if true padding added to left
 *     if omitted or false, padding added to right
 *
 * @return padded string or
 *     original string if length is >= padToLength
 */
Object.prototype.pad = function(padToLength, padChar, padLeft) {    

    // get the string value
    s = this.toString()

    // default padToLength to 0
    // if omitted, original string is returned
    padToLength = padToLength || 0;

    // default padChar to empty space
    padChar = padChar || ' ';


    // ignore padding if string too long
    if (s.length >= padToLength) {
        return s;
    }

    // create the pad of appropriate length
    var pad = Array(padToLength - s.length).join(padChar);

    // add pad to right or left side
    if (padLeft) {
        return pad  + s;        
    } else {
        return s + pad;
    }
};

Using the ECMAScript 6 method String#repeat and Arrow functions, a pad function is as simple as:

var leftPad = (s, c, n) => c.repeat(n - s.length) + s;
leftPad("foo", "0", 5); //returns "00foo"

jsfiddle

edit: suggestion from the comments:

const leftPad = (s, c, n) => n - s.length > 0 ? c.repeat(n - s.length) + s : s;

this way, it wont throw an error when s.lengthis greater than n

edit2: suggestion from the comments:

const leftPad = (s, c, n) =>{ s = s.toString(); c = c.toString(); return s.length > n ? s : c.repeat(n - s.length) + s; }

this way, you can use the function for strings and non-strings alike.

A variant of @Daniel LaFavers' answer.

var mask = function (background, foreground) {
  bg = (new String(background));
  fg = (new String(foreground));
  bgl = bg.length;
  fgl = fg.length;
  bgs = bg.substring(0, Math.max(0, bgl - fgl));
  fgs = fg.substring(Math.max(0, fgl - bgl));
  return bgs + fgs;
};

For example:

mask('00000', 11  );   // '00011'
mask('00011','00' );   // '00000'
mask( 2     , 3   );   // '3'
mask('0'    ,'111');   // '1'
mask('fork' ,'***');   // 'f***'
mask('_____','dog');   // '__dog'

http://www.webtoolkit.info/javascript_pad.html

/**
*
*  Javascript string pad
*  http://www.webtoolkit.info/
*
**/

var STR_PAD_LEFT = 1;
var STR_PAD_RIGHT = 2;
var STR_PAD_BOTH = 3;

function pad(str, len, pad, dir) {

    if (typeof(len) == "undefined") { var len = 0; }
    if (typeof(pad) == "undefined") { var pad = ' '; }
    if (typeof(dir) == "undefined") { var dir = STR_PAD_RIGHT; }

    if (len + 1 >= str.length) {

        switch (dir){

            case STR_PAD_LEFT:
                str = Array(len + 1 - str.length).join(pad) + str;
            break;

            case STR_PAD_BOTH:
                var padlen = len - str.length;
                var right = Math.ceil( padlen / 2 );
                var left = padlen - right;
                str = Array(left+1).join(pad) + str + Array(right+1).join(pad);
            break;

            default:
                str = str + Array(len + 1 - str.length).join(pad);
            break;

        } // switch

    }

    return str;

}

It's a lot more readable.

If you don't mind including a utility library, lodash library has _.pad, _.padLeft and _.padRight functions.

Here is a simple answer in basically one line of code.

var value = 35 // the numerical value
var x = 5 // the minimum length of the string

var padded = ("00000" + value).substr(-x);

Make sure the number of characters in you padding, zeros here, is at least as many as your intended minimum length. So really, to put it into one line, to get a result of "00035" in this case is:

var padded = ("00000" + 35).substr(-5);

All options included

function padding(stringToBePadded, paddingCharacter, totalLength, padLeftElseRight){
    //will pad any string provided in first argument, with padding character provide in 2nd argument and truncate to lenght provided in third argument, padding left if 4th argument true or undefined, right if false. 
    // i.e. padding("lode","x","10")  -->  "xxxxxxlode"
    // i.e. padding("lode","x","10",true)  -->  "xxxxxxlode"
    // i.e. padding("lode","x","10",false)  -->  "lodexxxxxx"
    // i.e. padding("12","0","5")  -->  "00012"
    {
        padLeftElseRight = typeof padLeftElseRight !== 'undefined' ? padLeftElseRight : true;
    }
    if (stringToBePadded.length > totalLength){
        // console.log("string too long to be padded");
        return stringToBePadded;
    }
    var paddingString = paddingCharacter.repeat(totalLength);//make long string of padding characters
    if ( padLeftElseRight){
        return String(paddingString+stringToBePadded).slice(-totalLength);
    }else{ 
        return String(stringToBePadded+paddingString).slice(0,totalLength); 
    }
}

Here is a JavaScript function that adds specified number of paddings with custom symble. the function takes three parameters.

    padMe --> string or number to left pad
    pads  --> number of pads
    padSymble --> custom symble, default is "0" 

    function leftPad(padMe, pads, padSymble) {
         if( typeof padMe === "undefined") {
             padMe = "";
         }
         if (typeof pads === "undefined") {
             pads = 0;
         }
         if (typeof padSymble === "undefined") {
             padSymble = "0";
         }

         var symble = "";
         var result = [];
         for(var i=0; i < pads ; i++) {
            symble += padSymble;
         }
        var length = symble.length - padMe.toString().length;
        result = symble.substring(0, length);
        return result.concat(padMe.toString());
    }

/* Here are some results:

> leftPad(1)
"1"
> leftPad(1, 4)
"0001"
> leftPad(1, 4, "0")
"0001"
> leftPad(1, 4, "@")
"@@@1"

*/

pad with default values

I noticed that i mostly need the padLeft for time conversion / number padding

so i wrote this function

function padL(a,b,c){//string/number,length=2,char=0
 return (new Array(b||2).join(c||0)+a).slice(-b)
}

This simple function supports Number or String as input

default pad is 2 chars

default char is 0

so i can simply write

padL(1);
// 01

if i add the second argument (pad width)

padL(1,3);
// 001

third parameter (pad char)

padL('zzz',10,'x');
// xxxxxxxzzz

EDIT @BananaAcid if you pass a undefined value or a 0 length string you get 0undefined..so:

as suggested

function padL(a,b,c){//string/number,length=2,char=0
 return (new Array((b||1)+1).join(c||0)+(a||'')).slice(-(b||2))
}

but this can also be achieved in a shorter way.

function padL(a,b,c){//string/number,length=2,char=0
 return (new Array(b||2).join(c||0)+(a||c||0)).slice(-b)
}

works also with:

padL(0)
padL(NaN)
padL('')
padL(undefined)
padL(false)

And if you want to be able to pad in both ways :

function pad(a,b,c,d){//string/number,length=2,char=0,0/false=Left-1/true=Right
return a=(a||c||0),c=new Array(b||2).join(c||0),d?(a+c).slice(0,b):(c+a).slice(-b)
}

which can be written in a shorter way without using slice.

function pad(a,b,c,d){
 return a=(a||c||0)+'',b=new Array((++b||3)-a.length).join(c||0),d?a+b:b+a
}
/*

Usage:

pad(
 input // (int or string) or undefined,NaN,false,empty string
       // default:0 or PadCharacter
 // optional
 ,PadLength // (int) default:2
 ,PadCharacter // (string or int) default:'0'
 ,PadDirection // (bolean) default:0 (padLeft) - (true or 1) is padRight 
)

*/

now if you try to pad 'averylongword' with 2 ... thats not my problem.

Said that i give you a tip.

Most of the time if you pad you do it for the same value N times.

Using any type of function inside a loop slows down the loop!!!

So if you just wanna pad left some numbers inside a long list don't use functions to do this simple thing.

use something like this:

var arrayOfNumbers=[1,2,3,4,5,6,7],
    paddedArray=[],
    len=arrayOfNumbers.length;
while(len--){
 paddedArray[len]=('0000'+arrayOfNumbers[len]).slice(-4);
}

if you don't know how the max padding size based on the numbers inside the array.

var arrayOfNumbers=[1,2,3,4,5,6,7,49095],
    paddedArray=[],
    len=arrayOfNumbers.length;

// search the highest number
var arrayMax=Function.prototype.apply.bind(Math.max,null),
// get that string length
padSize=(arrayMax(arrayOfNumbers)+'').length,
// create a Padding string
padStr=new Array(padSize).join(0);
// and after you have all this static values cached start the loop.
while(len--){
 paddedArray[len]=(padStr+arrayOfNumbers[len]).slice(-padSize);//substr(-padSize)
}
console.log(paddedArray);

/*
0: "00001"
1: "00002"
2: "00003"
3: "00004"
4: "00005"
5: "00006"
6: "00007"
7: "49095"
*/

Like PHP:

const STR_PAD_RIGHT = 1;
const STR_PAD_LEFT = 0;
const STR_PAD_BOTH = 2;

/**
 * @see http://php.net/str_pad
 * @param mixed input 
 * @param integer length 
 * @param string string 
 * @param integer type 
 * @return string
 */
function str_pad(input, length, string, type) {
    if (type === undefined || (type !== STR_PAD_LEFT && type !== STR_PAD_BOTH)) {
        type = STR_PAD_RIGHT
    }

    if (input.toString().length >= length) {
         return input;
    } else {
        if (type === STR_PAD_BOTH) {
            input = (string + input + string);
        } else if (type == STR_PAD_LEFT) {
            input = (string + input);
        } else {
            input = (input + string);
        }

        return str_pad(input.toString(), length, string, type);
    }
}

es7 is just drafts and proposals right now, but if you wanted to track compatibility with the spec, your pad functions need:

1. Multi-character pad support.
2. Don't truncate the input string
3. Pad defaults to space

From my polyfill library, but apply your own due diligence for prototype extensions.

// Tests
'hello'.lpad(4) === 'hello'
'hello'.rpad(4) === 'hello'
'hello'.lpad(10) === '     hello'
'hello'.rpad(10) === 'hello     '
'hello'.lpad(10, '1234') === '41234hello'
'hello'.rpad(10, '1234') === 'hello12341'

String.prototype.lpad || (String.prototype.lpad = function( length, pad )
{
    if( length < this.length ) return this;

    pad = pad || ' ';
    let str = this;

    while( str.length < length )
    {
        str = pad + str;
    }

    return str.substr( -length );
});

String.prototype.rpad || (String.prototype.rpad = function( length, pad )
{
    if( length < this.length ) return this;

    pad = pad || ' ';
    let str = this;

    while( str.length < length )
    {
        str += pad;
    }

    return str.substr( 0, length );
});

use repeat, it would be more simple.

    var padLeft=function(str, pad, fw){
        return fw>str.length ? pad.repeat(fw-str.length)+str : str;
    }

you can use it like: padeLeft('origin-str', '0', 20)

Taking up Samuel's ideas, upward here. And remember an old SQL script, I tried with this:

a=1234;
'0000'.slice(a.toString().length)+a;

It works in all the cases I could imagine:

a=     1 result  0001
a=    12 result  0012
a=   123 result  0123
a=  1234 result  1234
a= 12345 result 12345
a=  '12' result  0012

One liner if you want something compact:

String.prototype.pad = function(len, chr){_x000D_
        return((((new Array(len)).fill(chr)).join("") +this).substring(this.length));_x000D_
}

A friend asked about using a JavaScript function to pad left. It turned into a little bit of an endeavor between some of us in chat to code golf it. This was the result:

function l(p,t,v){
    v+="";return v.length>=t?v:l(p,t,p+v); 
}

It ensures that the value to be padded is a string, and then if it isn't the length of the total desired length it will pad it once and then recurse. Here is what it looks like with more logical naming and structure

function padLeft(pad, totalLength, value){
    value = value.toString();

    if( value.length >= totalLength ){
        return value;
    }else{
        return padLeft(pad, totalLength, pad + value);
    }
}

The example we were using was to ensure that numbers were padded with 0 to the left to make a max length of 6. Here is an example set:

function l(p,t,v){v+="";return v.length>=t?v:l(p,t,p+v);}_x000D_
_x000D_
var vals = [6451,123,466750];_x000D_
_x000D_
var pad = l(0,6,vals[0]);// pad with 0's, max length 6_x000D_
_x000D_
var pads = vals.map(function(i){ return l(0,6,i) });_x000D_
_x000D_
document.write(pads.join("<br />"));

ECMAScript 2017 adds a padStart method to the String prototype. This method will pad a string with spaces to a given length. This method also takes an optional string that will be used instead of spaces for padding.

'abc'.padStart(10);         // "       abc"
'abc'.padStart(10, "foo");  // "foofoofabc"
'abc'.padStart(6,"123465"); // "123abc"
'abc'.padStart(8, "0");     // "00000abc"
'abc'.padStart(1);          // "abc"

A padEnd method was also added that works in the same manner.

For browser compatibility (and a useful polyfill) see this link.