Regular expression to return text between brackets

Question

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

All I need is the contents inside the parenthesis.

Answer 1

If your problem is really that simple, you don't need a regex:

 s[s.find("(")+1:s.find(")")] 

Answer 2

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'" 

Answer 3

If you want to find all occurrences:

 >>> 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'] 

Answer 4

Based on tkerwin's answer if you 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 bracket and the last closing parenthesis to get (a+b)/(c+d) , because it searches for searches on the left of the line and stops in the first closing bracket.

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

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

Answer 5

 import re fancy = u'abcde(date=\'2/xc2/xb2\',time=\'/case/test.png\')' print re.compile( "\((.*)\)" ).search( fancy ).group( 1 )