python regular expression date formate

Question

Trying to write a RE to recognize date format mm/dd in Python

reg = "((1[0-2])|(0?[1-9]))/((1[0-9])|(2[0-9])|(3[0-1])|(0?[0-9]))"
match = re.findall(reg, text, re.IGNORECASE)
print match

For text = '4/13' it gives me

[('4', '4', '', '13', '13', '', '', '')]

but not

'4/13'

Thanks, Cheng

Possible duplicate of [Converting string into datetime](http://stackoverflow.com/questions/466345/converting-string-into-datetime) — , May 07 '12 at 15:04

score 3 · Answer 1 · answered May 07 '12 at 15:04

3

dont't use re.findall. use re.match:

reg = "((0?[1-9])|(1[0-2]))/((1[0-9])|(2[0-9])|(3[0-1])|(0?[0-9]))"
match = re.match(reg, text, re.IGNORECASE)
print match.group()

answered May 07 '12 at 15:04

mata

score 1 · Accepted Answer · answered May 07 '12 at 15:08

1

The other answers are more direct, but you could also add an extra pair of braces around your regex:

reg = "(((0?[1-9])|(1[0-2]))/((1[0-9])|(2[0-9])|(3[0-1])|(0?[0-9])))"

Now findall will give you:

[('4/13', '4', '4', '', '13', '13', '', '', '')]

You can now extract '4/13' from above.

answered May 07 '12 at 15:08

mohit6up

by the way, how do I get rid of the 4s and 13s? – cheng May 07 '12 at 15:10
You can't, using `findall` in this case. `findall` finds all patterns within pairs of braces. So you end up with the outer pattern `'4/13'` you are looking for along with some inner patterns like `'4'`, `'13'`... – mohit6up May 07 '12 at 15:21

2 Answers2