Is there a simple method to pull content between a regex? Assume I have the following sample text
SOME TEXT [SOME MORE TEXT] value="ssss" SOME MORE TEXT
My regex is:
compiledRegex = re.compile('\[.*\] value=("|\').*("|\')')
This will obviously return the entire [SOME MORE TEXT] value="ssss", however I only want ssss to be returned since that's what I'm looking for
I can obviously define a parser function but I feel as if python provides some simple pythonic way to do such a task
This is what capturing groups are designed to do.
compiledRegex = re.compile('\[.*\] value=(?:"|\')(.*)(?:"|\')')
matches = compiledRegex.match(sampleText)
capturedGroup = matches.group(1) # grab contents of first group
The ?: inside the old groups (the parentheses) means that the group is now a non-capturing group; that is, it won't be accessible as a group in the result. I converted them to keep the output simpler, but you can leave them as capturing groups if you prefer (but then you have to use matches.group(2) instead, since the first quote would be the first captured group).
Your original regex is too greedy: r'.*\]' won't stop at the first ']' and the second '.*' won't stop at '"'. To stop at c you could use [^c] or '.*?':
regex = re.compile(r"""\[[^]]*\] value=("|')(.*?)\1""")
m = regex.search("""SOME TEXT [SOME MORE TEXT] value="ssss" SOME MORE TEXT""")
print m.group(2)
value="foo'andvalue='bar"which you almost certainly don't want to do. You should use this instead:r'''\[.*\] value=("|')(.*?)\1'''. Note that using a triple-quoted string obviates the need to escape the "'". Also, it's a good idea to always use raw strings (e.g. r'foo' and r"bar") when dealing with regular expressions in Python.