JavaScript regex, different lengths but not range

Question

I'm interested in a concise way to specify different allowable lengths for a string of numbers. I want to allow strings of length 2, 4, 6, 8, and 10, but nothing in between. The following works just fine, but it is a bit long-winded:

var regex = /^([0-9]){2}|([0-9]){4}|([0-9]){6}|([0-9]){8}|([0-9]){10}$/;

Is there a shorter, less brute force way I can do this?

Thanks!

I think the alternations are the longest parts. The answer below (user3218114) looks like the best thing you can do with regex. Otherwise use regex to verify it's all numbers and then use something string.length() and check the value of that — skamazin
– skamazin, Commented Aug 5, 2014 at 17:52

Braj · Accepted Answer · 2014-08-05 17:55:53Z

3

I want to allow strings of length 2, 4, 6, 8, and 10, but nothing in between

You can try. A shorter version

^([0-9]{2}){1,5}$

DEMO

OR in simple term

^([0-9][0-9]){1,5}$

Enclose the whole regex inside parenthesis (...) for capturing groups.

Regex explanation:

  ^                        the beginning of the string
  (                        group and capture to \1 (between 1 and 5 times):
    [0-9]{2}                 any character of: '0' to '9' (2 times)
  ){1,5}                   end of \1
  $                        the end of the string

edited Aug 5, 2014 at 17:55

answered Aug 5, 2014 at 17:50

Braj

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7 Comments

T.J. Crowder Over a year ago

And you can go even further by throwing a \d in: ^((\d{2})){1,5}$ Fantastic lateral thinking here, nice one!

T.J. Crowder Over a year ago

I think your capture groups are in the wrong place, though: ^((?:\d{2}){1,5})$ (one capture group for the whole thing).

Federico Piazza Over a year ago

@user3218114 Even shorter changing [0-9] by \d and no need to double parenthesis. ^(\d{2}){1,5}$ would work too.

Braj Over a year ago

@T.J.Crowder It's just for match to validate not for grouping.

jrad Over a year ago

Ah very clever, this achieves exactly what I wanted. Thanks! Also thanks for the regex101.com link. I had no idea that website existed. P.S. I'll accept the answer when stackoverflow lets me.

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