How can I pad a value with leading zeros?

Question

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").

Answer 1

Since ECMAScript 2017 we have padStart:

const padded = (.1 + "").padStart(6, "0");
console.log(`-${padded}`);

Before ECMAScript 2017

With toLocaleString:

var n=-0.1;
var res = n.toLocaleString('en', {minimumIntegerDigits:4,minimumFractionDigits:2,useGrouping:false});
console.log(res);

Answer 2

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;
}

Answer 3

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? :)

Answer 4

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);
};

Answer 5

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

Answer 6

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;
};

Answer 7

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 ;)

Answer 8

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"

Answer 9

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"

Answer 10

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"

Answer 11

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/

Answer 12

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.

Answer 13

Yet another version :

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

Answer 14

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

Answer 15

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

Answer 16

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.

Answer 17

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 hexadecimal 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.

Answer 18

Use:

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.

Answer 19

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 sadly 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. After some time of banging my head for not having the right tool on JavaScript (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.

It does the same ("almost the same"; read on for detail) that the dream PHP function does, 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') + '.');

Answer 20

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;
}

Answer 21

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.

Answer 22

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

Answer 23

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")

Answer 24

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.

Answer 25

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";

Answer 26

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);
    }
}

Answer 27

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('.','');  
}

Answer 28

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);
}

Answer 29

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'

Answer 30

I use this snippet to get a five-digits representation:

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

Answer 31

I was here looking for a standard and had the same idea as Paul and Jonathan... Theirs are super cute, but 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.', '');
}

This 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'

Answer 32

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.

Answer 33

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

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

let n = 1
var zerofilled = ('0000'+n).slice(-4);
console.log(zerofilled)

Answer 34

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

Answer 35

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;
}

Answer 36

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'

Answer 37

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));
    }

Answer 38

Don't reinvent the wheel; use underscore string:

jsFiddle

var numToPad = '5';

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

Answer 39

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 ;)

Answer 40

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

Answer 41

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

Answer 42

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;
}

Answer 43

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;
}

Answer 44

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 + '';
}

Answer 45

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

Answer 46

This is the ES6 solution.

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

Answer 47

If you use Lodash.

var n = 1;

alert( _.padLeft(n, 2, 0) ); // 01

n = 10;

alert( _.padLeft(n, 2, 0) ); // 10

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

Answer 48

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

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

Answer 49

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 Mărășteanu 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.

Answer 50

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'