Python Regex, Finding what is in between two substrings

Question

The regex I have right now is:

string="hello (fdd()()()()(()(())()))"
re.match("%s\s*\((.*?)\)$"%re.escape("hello"), string)

And it will work well, the goal of it is to get whatever is inside the brackets. In this case: "fdd()()()()(()(())())"

I would like to make an alteration, the regex should work with this test case

"hello (hihihi(((())))hihi) { "

There is a curly-brace at the end of it. There should always be a curly-brace at the end of the given string therefore the first test case that I showed you will not work anymore (with the new regex that I want).

Try looking at it like this

hello[any amount of space]([get WHATEVER is inside here])[any amount of space]{[any amount of space]

I think that using the dollar sign is what is causing me problems. I am obviously not too familiar with using regex, so if anyone can help me that would be great. I am open to any solution including but not limited to other python modules, built in python string features, ect.

Thanks for your help,

HINT: If you think the $ is causing you problems, have you tried removing the $ (and the ?)? — mVChr
– mVChr, Commented Jul 6, 2012 at 21:52

BrtH · Accepted Answer · 2012-07-06 21:56:22Z

1

I think you could just use "hello\s*$(.+)$\s*{"

import re

text = "hello (hihihi(((())))hihi) { "
print re.match(r'hello\s*\((.+)\)\s*{', text).group(1)

gives me

hihihi(((())))hihi

You don't need the ? and $.

answered Jul 6, 2012 at 21:56

BrtH

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Andrew Clark · Accepted Answer · 2012-07-06 21:57:06Z

1

Try changing your regex to the following:

r"%s\s*\((.*?)\)\s*{" % re.escape("hello")

The only difference here is that the $ was replaced by \s*{, which means any amount of whitespace followed by a {.

Example:

>>> s = "hello (hihihi(((())))hihi) { "
>>> print re.match(r"%s\s*\((.*?)\)\s*{" % re.escape("hello"), s).group(1)
hihihi(((())))hihi

answered Jul 6, 2012 at 21:57

Andrew Clark

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