I saw this beautiful script to add thousands separator to js numbers:

function thousandSeparator(n, sep)
{
    var sRegExp = new RegExp('(-?[0-9]+)([0-9]{3})'),
        sValue = n + '';
    if(sep === undefined)
    {
        sep = ',';
    }
    while(sRegExp.test(sValue))
    {
        sValue = sValue.replace(sRegExp, '$1' + sep + '$2');
    }
    return sValue;
}

usage :

thousandSeparator(5000000.125, '\,') //"5,000,000.125"

However I'm having a trouble accepting the while loop.

I was thinking to change the regex to : '(-?[0-9]+)([0-9]{3})*' asterisk...

but now , how can I apply the replace statement ?

now I will have $1 and $2..$n

how can I enhance the replace func?

p.s. the code is taken from here /

I saw this beautiful script to add thousands separator to js numbers:

function thousandSeparator(n, sep)
{
    var sRegExp = new RegExp('(-?[0-9]+)([0-9]{3})'),
        sValue = n + '';
    if(sep === undefined)
    {
        sep = ',';
    }
    while(sRegExp.test(sValue))
    {
        sValue = sValue.replace(sRegExp, '$1' + sep + '$2');
    }
    return sValue;
}

usage :

thousandSeparator(5000000.125, '\,') //"5,000,000.125"

However I'm having a trouble accepting the while loop.

I was thinking to change the regex to : '(-?[0-9]+)([0-9]{3})*' asterisk...

but now , how can I apply the replace statement ?

now I will have $1 and $2..$n

how can I enhance the replace func?

p.s. the code is taken from here http://www.grumelo.com/2009/04/06/thousand-separator-in-javascript/

• javascript
• regex
• performance
• separator

• 1 Btw, it fails on 5000000.125678 -> 5,000,000.125,678 – zerkms Commented May 16, 2012 at 5:05
• @zerkms yep you right. grumelo.com/2009/04/06/thousand-separator-in-javascript – Royi Namir Commented May 16, 2012 at 5:06
• Check out this question - there's a link to an example of how to do what you're trying. – Rob I Commented May 16, 2012 at 5:07
• @RobI again , it still has a while loop. i think a better solution can be found....( or not) ...:) – Royi Namir Commented May 16, 2012 at 5:09
• @Cylian i think we need positive look ahead (-?[0-9]+)([0-9]{3})(?=\.) but it not working... i mean - only replace numbers which in their right theres a DOT. – Royi Namir Commented May 16, 2012 at 5:18

5 Answers 5

There is no need to use replace, you can just add toLocaleString instead:

console.log((5000000.125).toLocaleString('en'));

More information: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Number/toLocaleString

Your assumption

now i will have $1 and $2..$n

is wrong. You have two groups, because you have two sets of brackets.

    (-?[0-9]+)([0-9]{3})*
1.  ^^^^^^^^^^
2.            ^^^^^^^^^^

And then you repeat the second group. If it matches the second time, it overwrites the result of the first match, when it matches the third time, it overwrites ...

That means when matching is complete, $2 contains the value of the last match of that group.

First approach

(\d)(?=(?:[0-9]{3})+\b)

and replace with

$1,

See it on Regexr

It has the flaw that it does insert the comma also on the right of the dot. (I am working on it.)

Second approach

(\d)(?:(?=\d+(?=[^\d.]))(?=(?:[0-9]{3})+\b)|(?=\d+(?=\.))(?=(?:[0-9]{3})+(?=\.)))

and replace with

$1,

See it on Regexr

So now its getting a bit more complicated.

(\d)                   # Match a digit (will be reinserted)
(?:
    (?=\d+(?=[^\d.]))  # Use this alternative if there is no fractional part in the digit
    (?=(?:\d{3})+      # Check that there are always multiples of 3 digits ahead
    \b)                # Till a word boundary
    |                  # OR
    (?=\d+(?=\.))      # There is a fractional part
    (?=(?:\d{3})+      # Check that there are always multiples of 3 digits ahead
    (?=\.))            # Till a dot
)

Problem: does also match the fractional part if there is not the end of the string following.

Here is an ugly script to contrast your beautiful script.

10000000.0001 .toString().split('').reverse().join('')
.replace(/(\d{3}(?!.*\.|$))/g, '$1,').split('').reverse().join('')

Since we don't have lookbehinds, we can cheat by reversing the string and using lookaheads instead.

Here it is again in a more palatable form.

function thousandSeparator(n, sep) {

    function reverse(text) {
        return text.split('').reverse().join('');
    }

    var rx = /(\d{3}(?!.*\.|$))/g;

    if (!sep) {
        sep = ',';
    }

    return reverse(reverse(n.toString()).replace(rx, '$1' + sep));

}

How about this one:

result = "1235423.125".replace(/\B(?=(\d{3})+(?!\d))/g, ',') //1,235,423.125

Try this one:

result = subject.replace(/([0-9]+?)([0-9]{3})(?=.*?\.|$)/mg, "$1,$2");

Test here