[python] Regular expression to return text between parenthesis

u'abcde(date=\'2/xc2/xb2\',time=\'/case/test.png\')'

All I need is the contents inside the parenthesis.

import re

fancy = u'abcde(date=\'2/xc2/xb2\',time=\'/case/test.png\')'

print re.compile( "\((.*)\)" ).search( fancy ).group( 1 )

No need to use regex .... Just use list slicing ...

string="(tidtkdgkxkxlgxlhxl) ¥£%#_¥#_¥#_¥#"
print(string[string.find("(")+1:string.find(")")])

Building on tkerwin's answer, if you happen to have nested parentheses like in

st = "sum((a+b)/(c+d))"

his answer will not work if you need to take everything between the first opening parenthesis and the last closing parenthesis to get (a+b)/(c+d), because find searches from the left of the string, and would stop at the first closing parenthesis.

To fix that, you need to use rfind for the second part of the operation, so it would become

st[st.find("(")+1:st.rfind(")")]

If you want to find all occurences:

>>> re.findall('\(.*?\)',s)
[u"(date='2/xc2/xb2',time='/case/test.png')", u'(eee)']

>>> re.findall('\((.*?)\)',s)
[u"date='2/xc2/xb2',time='/case/test.png'", u'eee']

Use re.search(r'\((.*?)\)',s).group(1):

>>> import re
>>> s = u'abcde(date=\'2/xc2/xb2\',time=\'/case/test.png\')'
>>> re.search(r'\((.*?)\)',s).group(1)
u"date='2/xc2/xb2',time='/case/test.png'"

contents_re = re.match(r'[^\(]*\((?P<contents>[^\(]+)\)', data)
if contents_re:
    print(contents_re.groupdict()['contents'])