CODEWITHSUNDEEP
regex
hhanzo1
hhanzo1

Reputation: 647

Regex match 1 or 2 digits without error: look-behind requires fixed-width pattern?

How to capture 1 or 2 digits in a regex?

Text

IF-MIB::ifDescr.1 = 1/1/g1, Office to DMZ
IF-MIB::ifDescr.10 = 1/1/g10, Office to DMZ

Regex

This works for 1 digit only. ie. 1/1/g1 but does not work for 2 digits. ie. 1/1/1g10.

(?P<ifDescr>(?<=ifDescr.\d = ).*)

This works for 2 digits in the regex tester but not in python3 in centos.

(?P<ifDescr>(?<=ifDescr.\d|\d = ).*)

I get the following error:

re.error: look-behind requires fixed-width pattern

Upvotes: 0

Views: 192

Answers (1)

Hao Wu
Hao Wu

Reputation: 20724

Although you can't make a lookbehind non-fixed width, but you can group two lookbehinds with an or condition:

(?:(?<=ifDescr\.\d = )|(?<=ifDescr\.\d\d = ))(?P<ifDescr>.*)
  • (?:...) a non-capture group
  • (?<=ifDescr\.\d = ) lookbehind, match ifDescr. with one digit
  • | or
  • (?<=ifDescr\.\d\d = ) lookbehind, match ifDescr. with two digits
  • (?P<ifDescr>.*) match whatever after it, and set a group back reference called ifDescr

check the proof

But this works for limited combinations. If you're not sure how many digits there are, you should use group captures, and then get the result from group ifDescr

(?:ifDescr\.\d+ = )(?P<ifDescr>.*)

Upvotes: 0

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