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.
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+ timesSee 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 rightSee another regex demo
I think an alternative approach could be:
import re
pattern = re.compile("(\[[^\]]*\]\.\[[^\]]*\])")
print(pattern.findall(sss))
OUTPUT
['[Data Source].[Week]']
