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

Question

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.

Answer 1

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

Answer 2

I think an alternative approach could be:

import re

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

print(pattern.findall(sss))

OUTPUT

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