Regular expression to match 12345

Question

Is there a regex to match a string of increasing contiguous numbers, e.g. 123, 56789, etc? I don't think it can be in regex but worth checking with folks here.

Answer 1

^(?:0(?=1|$))?(?:1(?=2|$))?(?:2(?=3|$))?(?:3(?=4|$))?(?:4(?=5|$))?(?:5(?=6|$))?(?:6(?=7|$))?(?:7(?=8|$))?(?:8(?=9|$))?(?:9$)?$

Result: http://rubular.com/r/JfJJ6ntEQG

Answer 2

^(1|^)((2|^)((3|^)((4|^)((5|^)((6|^)((7|^)((8|^)((9|^))?)?)?)?)?)?)?)?$

Python demonstration:

>>> import re
>>> good  = ['1', '12', '123', '23', '3', '4', '45', '456', '56', '6', '7', '78', '789', '89', '9', '123456789']
>>> bad   = ['a', '11', '13', '1345', '2459', '321', '641', '1 2', '222233334444']
>>> tests = good + bad
>>>
>>> regex = '^(1|^)((2|^)((3|^)((4|^)((5|^)((6|^)((7|^)((8|^)((9|^))?)?)?)?)?)?)?)?$'
>>> for test in tests:
...   print '%s: %s' % (re.match(regex, test) and 'Passed' or 'Failed', test)
... 
Passed: 1
Passed: 12
Passed: 123
Passed: 23
Passed: 3
Passed: 4
Passed: 45
Passed: 456
Passed: 56
Passed: 6
Passed: 7
Passed: 78
Passed: 789
Passed: 89
Passed: 9
Passed: 123456789
Failed: a
Failed: 11
Failed: 13
Failed: 1345
Failed: 2459
Failed: 321
Failed: 641
Failed: 1 2
Failed: 222233334444