Python Regex to match optionally double-quoted string

Question

I want to match an optionally double-quoted string with regular expression using Python regex module re

The expression should give the following results:

"Assets". => Should Match

Assets. => Should Match

"Assets. => Shouldn't Match

Assets". => Shouldn't Match

I tried to achieve this using back reference in regular expression :

("?)Assets\1 

However, it matches even if there is no matching end quote. "Assets. -> neglects initial quote ", and matches the rest of the word.

What would be right expression for this ?

Answer 1

You can use the following pattern. Note that it basically lists the two separate cases because parentheses are notoriously not regular, but context-sensitive and, thus, difficult to handle with regular expressions:

>>> p = re.compile(r'^(?:"[^"]+"|[^"]+)$')
>>> bool(p.match('"assets"'))
True
>>> bool(p.match('"assets'))
False
>>> bool(p.match('assets'))
True

This also assumes that are no chars before or after the string that is being matched.

Answer 2

You regexp pattern is almost correct. You just have to make sure there are no quotes before and after your pattern. So use the pattern r'(?<!")("?)Assets\\1(?!")

>>> words = ['"Assets"', 'Assets', '"Assets', 'Assets"']
>>> ptrn = re.compile(r'(?<!")("?)Assets\1(?!")')
>>> [bool(ptrn.match(word)) for word in words]
[True, True, False, False]