Conditional(?) regex for numbers

user_17264570 0
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i was trying to create a regex that could match numbers within brackets or not, for example:

(1.000.000,00) //match
(1.000,00)     //match
(100,00)       //match
(10)           //match
(1)            //match

(2.000.000,00  //dont't match
(2.000,00      //dont't match
(200,00        //dont't match
(20            //dont't match
(2             //dont't match

3.000.000,00)  //dont't match
3.000,00)      //dont't match
300,00)        //dont't match
30)            //dont't match
3)             //dont't match

4.000.000,00   //should match
4.000,00       //should match
400,00         //should match
40             //should match
4              //should match

I need to match only numbers(in brackets or not), but only if they have all brackets (2) or none(0)

At the moment this is what i came up with: \((\d+[\.,]?)+\d*\), it matches the --match and doesn't match the --don't match but should match also the --should match

I've added the javascript tag because i'm using this regex in js and not all of the regex tokens work in the js regex constructor

I'm posting also a regex101 link

javascript regex
3 Answers

If supported, you can usea negative lookbehind to match either with or without parenthesis:

\(\d+(?:[.,]\d+)*\)|(?<!\S)\d+(?:[.,]\d+)*(?!\S)
  • \( Match (
  • \d+(?:[.,]\d+)* Match 1+ digits and optionally repeat matching . or , and again 1+ digits
  • \) Match )
  • | Or
  • (?<!\S) Negative lookbehind, assert a word boundary to the left
  • \d+(?:[.,]\d+)* Match 1+ digits and optionally repeat matching . or , and again 1+ digits
  • (?!\S) Negative lookahead, assert a whitespace boundary to the right

Regex demo

Another option could be matching optional parenthesis at both sides, and only keep the ones that have either an opening and closing parenthesis, or none.

const regex = /\(?\d+(?:[.,]?\d+)*\)?/
const strings = ["(1.000.000,00)", "(1.000,00)", "(100,00)", "(10)", "(1)", "(2.000.000,00", "(2.000,00", "(200,00", "(20", "(2", "3.000.000,00)", "3.000,00)", "300,00)", "30)", "3)", "4.000.000,00", "4.000,00", "400,00", "40", "4"];

strings.forEach(s => {

  const m = s.match(regex);
  const firstChar = s.charAt(0);
  const lastChar = s.charAt(s.length - 1);

  if (
    m &&
    (firstChar !== '(' && lastChar !== ')') ||
    firstChar === '(' && lastChar === ')'
  ) {
    console.log(s)
  }
});

user_54249880

Edit: this is broken. It will match numbers like (7 as it only matches the number and ignores the parentheses in that case.

Kept here for future reference.

It's usually easier to do regex in multiple passes, but here goes:

/(\((\d+[\.,]?)+\d*\))|(\d+[\.,]?\d*)/gm

You can test it on https://regex101.com/. Usually it's better to process something in multiple passes as you can see the regex becomes even more unreadable. I took your regex and just split it into two regexes: one that requires the parentheses and one that doesn't, then combined them with the or operator. Note that this regex will allow things like "123.3,5.7" as one number, and the capturing groups will be nasty.

user_78183490

If you don't want to repeat the part matching numbers (which in this case is short, so maybe an exception to the DRY rule is warranted), you can reach for \(?((\d+[\.,]?)+\d*)\)?(?<=\(\1\)|(^|[^(\d.,])\1(?=($|[^\d.,)]))).

user_37777170
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