Regex string between square brackets only if '.' is within string

Tim Pickup 0
Viewed 40

I'm trying to detect the text between two square brackets in Python however I only want the result where there is a "." within it.

I currently have [(.*?] as my regex, using the following example:

String To Search: CASE[Data Source].[Week] = 'THIS WEEK'

Result: Data Source, Week

However I need the whole string as [Data Source].[Week], (square brackets included, only if there is a '.' in the middle of the string). There could also be multiple instances where it matches.

regex python
2 Answers

You might write a pattern matching [...] and then repeat 1 or more times a . and again [...]

\[[^][]*](?:\.\[[^][]*])+

Explanation

  • \[[^][]*] Match from [...] using a negated character class
  • (?: Non capture group to repeat as a whole part
    • \.\[[^][]*] Match a dot and again [...]
  • )+ Close the non capture group and repeat 1+ times

See a regex demo.

To get multiple matches, you can use re.findall

import re

pattern = r"\[[^][]*](?:\.\[[^][]*])+"

s = ("CASE[Data Source].[Week] = 'THIS WEEK'\n"
            "CASE[Data Source].[Week] = 'THIS WEEK'")

print(re.findall(pattern, s))

Output

['[Data Source].[Week]', '[Data Source].[Week]']

If you also want the values of between square brackets when there is not dot, you can use an alternation with lookaround assertions:

\[[^][]*](?:\.\[[^][]*])+|(?<=\[)[^][]*(?=])

Explanation

  • \[[^][]*](?:\.\[[^][]*])+ The same as the previous pattern
  • | Or
  • (?<=\[)[^][]*(?=]) Match [...] asserting [ to the left and ] to the right

See another regex demo

The fourth bird0

I think an alternative approach could be:

import re

pattern = re.compile("(\[[^\]]*\]\.\[[^\]]*\])")

print(pattern.findall(sss))

OUTPUT

['[Data Source].[Week]']
nikeros0
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