Achieving format x xxx,xx using replace

Question

I am trying to get a replace method to format the string to something like:

100 000,10

If the value written was:

100j00 0,1 0

Always two decimals and for every three integers to insert a space(counting thousands). So far I have not managed that outcome.

const parse = (input) => {

    const numberString = input ? input.toString().replace(/[^0-9]/g, '') : '';
    return parseInt(numberString);
};

const NumberFormatFilter = () => {

    return (input) => {

        const number = parse(input);

        return isNaN(number) ? '' :
            number
                .toLocaleString()
                .replace(/\./g, ' ')
                .replace(/,/g, ' ');
    };
};

This will get me to a format x xxx(1 000) but I am having problems with making it into a format with x xxx,xx(1 000,20)

https://stackoverflow.com/questions/2901102/how-to-print-a-number-with-commas-as-thousands-separators-in-javascript same idea, just a space instead of a comma — epascarello, Sep 04 '19 at 13:57
@epascarello No that is not a typo, I want to be able to filter characters out. — Yooric, Sep 04 '19 at 13:59
It is kind of same but different. That post wants "1,234,567" from "1234567" but I want "12345,67" to "12 345,67" — Yooric, Sep 04 '19 at 14:02
Yeah, it is the same thing, use the different separators..... — epascarello, Sep 04 '19 at 14:09

epascarello · Accepted Answer · 2019-09-04T14:43:17.283

So clean up the string to remove anything that is not a number or comma. And than split it and process it. Basic idea is below.

function numberWithSep(x) {
  var parts = x.toString().split(",");
  //deal with the whole number format
  parts[0] = parts[0].replace(/\B(?=(\d{3})+(?!\d))/g, " ");

  //deal with the decimal format (no rounding here)
  parts[1] = ((parts[1] || '') + "00").substr(0,2)

  // If you want rounding
  //parts[1] = Number('.' + (parts[1] || '0')).toFixed(2).substr(2,2)
  
  
  return parts.join(",");
}

function myFormat (str) {
  const cleaned = str.replace(/[^\d,]/g,'')
  return numberWithSep(cleaned)
}

const tests = [
  '100 j00 0,1 0',
  '1000',
  '1 0 0 0',
  '1000,01',
  '1234,569',
  '12345678901234567890,12',
]

tests.forEach( function (str) {
  console.log(myFormat(str))
})

Other option is to just format the number with toLocaleString.

function myFormat (str) {
  const cleaned = str.replace(/[^\d,]/g,'').replace(",",".")
  return Number(cleaned).toLocaleString('fr', { minimumFractionDigits: 2, maximumFractionDigits: 2 })
}

const tests = [
  '100 j00 0,1 0',
  '1000',
  '1 0 0 0',
  '1000,01',
  '1234,569',
  '12345678901234567890,12',
]

tests.forEach( function (str) {
  console.log(myFormat(str))
})

Looks good =D But it will not work on 12,12,312. But I can solve that by counting amounts of "," Thanks alot =) — Yooric, Sep 10 '19 at 10:54
well `12,12,312` is not really a valid input unless your style of math somehow has two decimal places. — epascarello, Sep 10 '19 at 12:34
Forgot to mark as answer =) Thanks alot! (New to actually posting on stackorverflow) — Yooric, Sep 17 '19 at 06:47

Achieving format x xxx,xx using replace

1 Answers1