How can I format a number with commas as thousands separators?

I am trying to print an integer in JavaScript with commas as thousands separators. For example, I want to show the number 1234567 as "1,234,567". How would I go about doing this?

Here is how I am doing it:

function numberWithCommas(x) {
x = x.toString();
var pattern = /(-?\d+)(\d{3})/;
while (pattern.test(x))
x = x.replace(pattern, "$1,$2");
return x;
}

console.log(numberWithCommas(1000))

Is there a simpler or more elegant way to do it? It would be nice if it works with floats also, but that is not necessary. It does not need to be locale-specific to decide between periods and commas.

I used the idea from Kerry's answer, but I simplified it since I was just looking for something simple for my specific purpose. Here is what I have:

function numberWithCommas(x) {
    return x.toString().replace(/\B(?=(\d{3})+(?!\d))/g, ",");
}

function numberWithCommas(x) {
return x.toString().replace(/\B(?"} ${x} => ${result});
return pass;
}

let failures = 0;
failures += !test(0, "0");
failures += !test(100, "100");
failures += !test(1000, "1,000");
failures += !test(10000, "10,000");
failures += !test(100000, "100,000");
failures += !test(1000000, "1,000,000");
failures += !test(10000000, "10,000,000");
if (failures) {
console.log(${failures} test(s) failed);
} else {
console.log("All tests passed");
}
.as-console-wrapper {
max-height: 100% !important;
}

The regex uses two lookahead assertions:

• a positive one to look for any point in the string that has a multiple of 3 digits in a row after it,

• a negative assertion to make sure that point only has exactly a multiple of 3 digits. The replacement expression puts a comma there.

For example, if you pass it 123456789.01, the positive assertion will match every spot to the left of the 7 (since 789 is a multiple of 3 digits, 678 is a multiple of 3 digits, 567, etc.).

The negative assertion checks that the multiple of 3 digits does not have any digits after it. 789 has a period after it so it is exactly a multiple of 3 digits, so a comma goes there. 678 is a multiple of 3 digits, but it has a 9 after it, so those 3 digits are part of a group of 4, and a comma does not go there. Similarly for 567.

456789 is 6 digits, which is a multiple of 3, so a comma goes before that. 345678 is a multiple of 3, but it has a 9 after it, so no comma goes there. And so on. The \B keeps the regex from putting a comma at the beginning of the string.

neu-rah mentioned that this function adds commas in undesirable places if there are more than 3 digits after the decimal point. If this is a problem, you can use this function:

function numberWithCommas(x) {
    var parts = x.toString().split(".");
    parts[0] = parts[0].replace(/\B(?=(\d{3})+(?!\d))/g, ",");
    return parts.join(".");
}

function numberWithCommas(x) {
var parts = x.toString().split(".");
parts[0] = parts[0].replace(/\B(?=(\d{3})+(?!\d))/g, ",");
return parts.join(".");
}

function test(x, expect) {
const result = numberWithCommas(x);
const pass = result === expect;
console.log(${pass ? "✓" : "ERROR ====>"} ${x} => ${result});
return pass;
}

let failures = 0;
failures += !test(0 , "0");
failures += !test(0.123456 , "0.123456");
failures += !test(100 , "100");
failures += !test(100.123456 , "100.123456");
failures += !test(1000 , "1,000");
failures += !test(1000.123456 , "1,000.123456");
failures += !test(10000 , "10,000");
failures += !test(10000.123456 , "10,000.123456");
failures += !test(100000 , "100,000");
failures += !test(100000.123456 , "100,000.123456");
failures += !test(1000000 , "1,000,000");
failures += !test(1000000.123456 , "1,000,000.123456");
failures += !test(10000000 , "10,000,000");
failures += !test(10000000.123456, "10,000,000.123456");
if (failures) {
console.log(${failures} test(s) failed);
} else {
console.log("All tests passed");
}
.as-console-wrapper {
max-height: 100% !important;
}

T.J. Crowder pointed out that now that JavaScript has lookbehind (support info), it can be solved in the regular expression itself:

function numberWithCommas(x) {
    return x.toString().replace(/\B(?<!\.\d*)(?=(\d{3})+(?!\d))/g, ",");
}

function numberWithCommas(x) {
return x.toString().replace(/\B(?"} ${x} => ${result});
return pass;
}

let failures = 0;
failures += !test(0, "0");
failures += !test(0.123456, "0.123456");
failures += !test(100, "100");
failures += !test(100.123456, "100.123456");
failures += !test(1000, "1,000");
failures += !test(1000.123456, "1,000.123456");
failures += !test(10000, "10,000");
failures += !test(10000.123456, "10,000.123456");
failures += !test(100000, "100,000");
failures += !test(100000.123456, "100,000.123456");
failures += !test(1000000, "1,000,000");
failures += !test(1000000.123456, "1,000,000.123456");
failures += !test(10000000, "10,000,000");
failures += !test(10000000.123456, "10,000,000.123456");
if (failures) {
console.log(${failures} test(s) failed);
} else {
console.log("All tests passed");
}
.as-console-wrapper {
max-height: 100% !important;
}

(?<!\.\d*) is a negative lookbehind that says the match can't be preceded by a . followed by zero or more digits. The negative lookbehind is faster than the split and join solution (comparison), at least in V8.

Stack Overflow Q#2901102, score: 3954