Validating a numeric, 12-digit String containing a dash using regular expressions

Question

I need to validate that a given String contains up to 12 digits and exactly one dash.

Initial RegEx: ^[0-9]*-?[0-9]*$

Modified RegEx: ^([0-9]*-?[0-9]*){1,12}$

Example (should be valid): 12356978-9

The problem is that the first RegEx doesn't validate the length, and the second one doesn't work.

Note: Everything must be done in regex, not checking length using string.length()

It would be good if you give some example strings you want to compare with. — Chirag Parmar, Nov 18 '16 at 13:02
123465789-10 is valid, 123456789-1011 is invalid because of lenght,123465789 is invalid because of no dash in it — Abdul Majid Bajwa, Nov 18 '16 at 13:04
Something along `String s = ... ; return s.length()<=13 && s.matches("^\\d+-\\d+$");`? — Olivier Grégoire, Nov 18 '16 at 13:04
@Seth You changed the question. Nowhere it was said that no dashes was an acceptable solution! — Olivier Grégoire, Nov 18 '16 at 13:10
@OlivierGrégoire Noted. The "up to" part confused me, I thought it was meant as "up to 12 digits as well as up to one dash". Thanks for the fix! — Seth, Nov 19 '16 at 15:17

score 2 · Accepted Answer · edited May 23 '17 at 11:59

2

The ugly way:

^([0-9]-[0-9]{1,11}|[0-9]{2}-[0-9]{1,10}|[0-9]{3}-[0-9]{1,9}| ...)$

Using lookahead, combining two conditions:

^(?=\\d*-\\d*$)(?=.{1,13}$).*$

edited May 23 '17 at 11:59

Community

answered Nov 18 '16 at 13:03

Jiri Tousek

The second one should be the accepted answer. I'd +1 again if I could, but I already did :( – Olivier Grégoire Nov 19 '16 at 16:55

Olivier Grégoire · Answer 2 · 2016-11-18T13:11:49.783

2

If you can use extra conditions outside of the regex:

String s = ... ;
return s.length()<=13 && s.matches("^\\d+-\\d+$");

If a dash can start or end the string, you can use the following:

String s = ... ;
return s.length()<=13 && s.matches("^\\d*-\\d*$");

edited Nov 18 '16 at 13:11

answered Nov 18 '16 at 13:06

Olivier Grégoire

Any idea on how to implement in pure regex? – Mohammad Yusuf Nov 18 '16 at 13:13
1

Yes, you can then implement @JiriTousek 's solution. Not very readable, but will do the job. This doesn't mean that there is no solution, just that if there is, other than Jiri Tousek, I don't know it. – Olivier Grégoire Nov 18 '16 at 13:15

2 Answers2