Question

I would like to be able to match a string literal with the option of escaped quotations. For instance, I’d like to be able to search ‘this is a ‘test with escaped\’ values’ ok’ and have it properly recognize the backslash as an escape character. I’ve tried solutions like the following:

import re regexc = re.compile(r'\'(.*?)(?<!\\)\'') match = regexc.search(r''' Example: 'Foo \' Bar'  End. ''') print match.groups()  # I want ('Foo \' Bar') to be printed above 

After looking at this, there is a simple problem that the escape character being used, ‘\‘, can’t be escaped itself. I can’t figure out how to do that. I wanted a solution like the following, but negative lookbehind assertions need to be fixed length:

# ... re.compile(r'\'(.*?)(?<!\\(\\\\)*)\'') # ... 

Any regex gurus able to tackle this problem? Thanks.

Answer 1

I think this will work:

import re regexc = re.compile(r'(?:^|[^\\])'(([^\\']|\\'|\\\\)*)'')  def check(test, base, target):     match = regexc.search(base)     assert match is not None, test+': regex didn't match for '+base     assert match.group(1) == target, test+': '+target+' not found in '+base     print 'test %s passed'%test  check('Empty','''','') check('single escape1', r''' Example: 'Foo \' Bar'  End. ''',r'Foo \' Bar') check('single escape2', r''''\''''',r'\'') check('double escape',r''' Example2: 'Foo \\' End. ''',r'Foo \\') check('First quote escaped',r'not matched\''a'','a') check('First quote escaped beginning',r'\''a'','a') 

The regular expression r'(?:^|[^\\])'(([^\\']|\\'|\\\\)*)'' is forward matching only the things that we want inside the string:

1. Chars that aren’t backslash or quote.
2. Escaped quote
3. Escaped backslash

EDIT:

Add extra regex at front to check for first quote escaped.