Enhancing regex of thousands separator?

Question

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         


        

Answer 1

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

Answer 2

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));

}

Answer 3

How about this one:

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

Answer 4

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.

Answer 5

Try this one:

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

Test here