Regular expression non-capturing group character length

Question

I'm trying to generate a regular expression with the next pattern. A number, of a maximum of 16 digits, that can have or not a comma inside, never at the beginning, never at the end. I tried this:

^(?:\d+(?:,{1}\d+){1})$

The problem is that I can't count the result of a group {0,16}. This is a list of numbers that should fit the expression:

123,34
1,33333333
1222222233

Example numbers that shouldn't fit:

1111,1111,1111
,11111
11111,
11111111111111111111111111111,111111111111111 (more than 16 characters)

Answer 1

If your regex flavour supports lookahead assertions, you can use this:

^(?!(?:\d{17,}$|[,\d]{18,}$))(?:\d+(?:,\d+)?)$

See it here on Regexr

I removed the superfluous {1} and made the group with the fraction optional.

The negative lookahead assertion (?!(?:\d{17,}$|[,\d]{18,}$)) is checking your length requirement. It fails if it finds 17 or more digits till the end OR 18 or more digits and commas till the end. That I allow multiple commas in the character class here is not a problem, that there is only one comma is ensured by the following pattern.

Answer 2

You may check the length before that or using ^(?=[\d,]{1,16}$)(?:\d+(?:,\d+)?)$

That is a lookahead that checks the length before doing the real match.