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,$2"); return x; } console.log(numberWithCommas(1000))
Is there a simpler or more elegant way to do it? It would be nice if it works with floats also, but that is not necessary. It does not need to be locale-specific to decide between periods and commas.
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Answer
I used the idea from Kerry’s answer, but simplified it since I was just looking for something simple for my specific purpose. Here is what I have:
function numberWithCommas(x) { return x.toString().replace(/B(?=(d{3})+(?!d))/g, ","); }
function numberWithCommas(x) { return x.toString().replace(/B(?<!.d*)(?=(d{3})+(?!d))/g, ","); } function test(x, expect) { const result = numberWithCommas(x); const pass = result === expect; console.log(`${pass ? "✓" : "ERROR ====>"} ${x} => ${result}`); return pass; } let failures = 0; failures += !test(0, "0"); failures += !test(100, "100"); failures += !test(1000, "1,000"); failures += !test(10000, "10,000"); failures += !test(100000, "100,000"); failures += !test(1000000, "1,000,000"); failures += !test(10000000, "10,000,000"); if (failures) { console.log(`${failures} test(s) failed`); } else { console.log("All tests passed"); }
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The regex uses 2 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"); }
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@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(?<!.d*)(?=(d{3})+(?!d))/g, ","); } 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"); }
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(?<!.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.