Regex - find string which does not contain certain char

Question

I'm trying to find a way to write regex which match any 5 digit numbers in string, except those, which are followed by slash.

I can't figure out why this RE doesn't work:

r"\D(\d{%d})[^/]\D" % 5

EXAMPLE

'dsadasd894665' -> NO MATCH
'dsadsa78954,4' -> 78954 
'dsda78954/sdd' -> NO MATCH

Have you any idea?

Answer 1

(?<!\d)(\d{5})(?!\/|\d)

Try this.This Works.See demo.Your regex does not start or end with 0 width assertions.So it actually needs 1 charcter before and 2 ahead.

http://regex101.com/r/yA5iD9/16

Answer 2

You could try the below regex,

(?<!\d)\d{5}(?=[^\d/]|$)

DEMO

EXplanation:

• (?<!\\d) Match wouldn't be preceded by a digit.
• \\d{5} Exactly 5 digits.
• (?=[^\\d/]|$) Must be followed by a non-digit or / or end of the line anchor.

Code:

>>> import re
>>> s1 = "dsadasd894665"
>>> s2 = "dsadsa78954,4"
>>> s3 = "dsda78954/sdd"
>>> re.search(r'(?<!\d)\d{5}(?=[^\d/]|$)', s1)
>>> re.search(r'(?<!\d)\d{5}(?=[^\d/]|$)', s2)
<_sre.SRE_Match object at 0x7f0835855370>
>>> re.search(r'(?<!\d)\d{5}(?=[^\d/]|$)', s3)

Answer 3

The reason your regex is not working is due to the \\D s. I don't know why you put them in. Another problem is that [^/] will not match the end of the string, so use ?! instead.

The following works:

r"(\d{%d})(?!\/)" % 5

Answer 4

Since you want to find numbers that are on their own, ie not preceeded by another number, not followed by another number, and also not followed by a slash, we are going to use negative look-aheads and look-behinds to ensure this:

(?<!\d)\d{5}(?!/|\d)

This will match a \\d{5} if it is not preceeded by another \\d (ie no number is before it), and it is also not followed by another \\d or a / .

Using that on your examples yields the desired results:

>>> for example in ('dsadasd894665', 'dsadsa78954,4', 'dsda78954/sdd'):
    print(re.search(r'(?<!\d)\d{%d}(?!/|\d)' % 5, example))

None
<_sre.SRE_Match object; span=(6, 11), match='78954'>
None