Regular Expression that return matches specific strings in bracket and return its next and preceding string in brackets

Question

I need to find the string in brackets that matches some specific string and all values in the string. Right not I am getting values from a position where the string matches.

text = 'This is an sample string which have some information in brackets (info; matchingString, someotherString).'

regex= r"\(*?matchingString.*?\)"
matches = re.findall(regex, text)

From this I am getting result matchingString, someotherString) what I want is to get the string before the matching string as well. The result should be like this: (info; matchingString, someotherString) This regex works if the matching string is in the first string in brackets.

Answer 1

You can use

\([^()]*?matchingString[^)]*\)

See the regex demo. Due to the [^()]*?, the match will never overflow across other (...) substrings.

Regex details:

• \( - a ( char
• [^()]*? - zero or more chars other than ( and ) as few as possible
• matchingString - a hardcoded string
• [^)]* - zero or more chars other than )
• \) - a ) char.

See the Python demo:

import re
text = 'This is an sample string which have some information in brackets (info; matchingString, someotherString).'
regex= r"\([^()]*?matchingString[^)]*\)"
print( re.findall(regex, text) )
# => ['(info; matchingString, someotherString)']