Regex for managing escaped characters for items like string literals

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\\\' val

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.