I recently came up with this code while answering another StackOverflow question. Basically, on blur, this code will properly comma separate by thousands and leave the decimal at two digits (like how USD is written [7,745.56]).
I was wondering if there is more concise way of using regex to , separate and cut off excessive decimal places. I recently updated this post with my most recent attempt. Is there a better way of doing this with regex?
7456 -> 7,456
45345 -> 45,345
25.23523534 -> 25.23
3333.239 -> 3,333.23
234.99 -> 234.99
2300.99 -> 2,300.99
23123123123.22 -> 23,123,123,123.22
var result;
var str = []
reg = new RegExp(/(\d*(\d{2}\.)|\d{1,3})/, "gi");
reversed = "9515321312.2323432".split("").reverse().join("")
while (result = reg.exec(reversed)) {
str.push(result[2] ? result[2] : result[0])
}
console.log(str.join(",").split("").reverse().join("").replace(",.","."))As an alternative to the Regex, you could use the following approach
Number(num.toFixed(2)).toLocaleString('en-US')
or
num.toLocaleString('en-US', {maximumFractionDigits: 2})
You would still have the toFixed(2), but it's quite clean. toFixed(2) though won't floor the number like you want. Same with {maximumFractionDigits: 2} as the second parameter to toLocaleString as well.
var nums = [7456, 45345, 25.23523534, 3333.239, 234.99, 2300.99, 23123123123.22]
for (var num of nums)
console.log(num, '->', Number(num.toFixed(2)).toLocaleString('en-US') )Flooring the number like you showed is a bit tricky. Doing something like (num * 100 | 0) / 100 does not work. The calculation loses precision (e.g. .99 will become .98 in certain situations). (also |0 wouldn't work with larger numbers but even Math.floor() has the precision problem).
The solution would be to treat the numbers like strings.
function format(num) {
var num = num.toLocaleString('en-US')
var end = num.indexOf('.') < 0 ? num.length : num.indexOf('.') + 3
return num.substring(0, end)
}
var nums = [7456, 45345, 25.23523534, 3333.239, 234.99, 2300.99, 23123123123.22]
for (var num of nums) console.log(num, '->', format(num))
function format(num) {
var num = num.toLocaleString('en-US')
var end = num.indexOf('.') < 0 ? num.length : num.indexOf('.') + 3
return num.substring(0, end)
}(when changing to another format than 'en-US' pay attention to the . in numbers as some languages use a , as fractal separator)
For Compatibility, according to CanIUse toLocaleString('en-US') is
supported in effectively all browsers (since IE6+, Firefox 2+, Chrome 1+ etc)
If you really insist on doing this purely in regex (and truncate instead of round the fractional digits), the only solution I can think of is to use a replacement function as the second argument to .replace():
('' + num).replace(
/(\d)(?=(?:\d{3})+(?:\.|$))|(\.\d\d?)\d*$/g,
function(m, s1, s2){
return s2 || (s1 + ',');
}
);
This makes all your test cases pass:
function format(num){
return ('' + num).replace(
/(\d)(?=(?:\d{3})+(?:\.|$))|(\.\d\d?)\d*$/g,
function(m, s1, s2){
return s2 || (s1 + ',');
}
);
}
test(7456, "7,456");
test(45345, "45,345");
test(25.23523534, "25.23"); //truncated, not rounded
test(3333.239, "3,333.23"); //truncated, not rounded
test(234.99, "234.99");
test(2300.99, "2,300.99");
test(23123123123.22, "23,123,123,123.22");
function test(num, expected){
var actual = format(num);
console.log(num + ' -> ' + expected + ' => ' + actual + ': ' +
(actual === expected ? 'passed' : 'failed')
);
}Try:
var n = 5812090285.2817481974897;
n = n.toFixed(2).replace(/(\d)(?=(\d{3})+\.)/g, '$1,');
console.log(n);
Outputs:
5,812,090,285.28
Note: .toFixed(2) returns a string. So in order to simplify this further you must add a way to turn n into a string before executing your regex. For example:
n.toString.replace(/(\d)(?=(\d{3})+\.)/g, '$1,'); //ofc with the additional regex
Although you would think it wouldn't matter in javascript, it apparently does in this situation. So I dont know how much 'less' messy it would be to not use.