Regex for string matching with special escaping rule

Question

I'm trying to match special kind of string literals with some funky escaping rules.

The general form looks like this:

"some string"

Which are simple to match using a pattern such as "(.*?)"

However you can escape quotes by doubling them, such as:

"hello "" there" becomes hello " there
"hello """" there" becomes hello "" there

And this is where my regex skills fail me. How can I match strings like this?

Oh, and I'm using python 3.1.

I think that this way of formatting is better. Feel free to rollback if you don't think so. — Jerry
– Jerry, Commented Jun 3, 2013 at 18:36
Although, I'm still not sure what you are trying to do. You have input strings as "hello "" there" and "hello """" there" and want the output to be hello " there and hello "" there respectively or? — Jerry
– Jerry, Commented Jun 3, 2013 at 18:41
The regex doesn't have to unescape the string, I can do that later, however I need to match the entire string. My current pattern stops matching when hitting the double " while it should continue until the first " which isn't doubled. — monoceres
– monoceres, Commented Jun 3, 2013 at 18:45
what would you expect in case of """? Can you match the string first and then replace the quotes? — Joanna Derks
– Joanna Derks, Commented Jun 3, 2013 at 18:46

morningstar · Accepted Answer · 2013-06-03 18:48:23Z

3

regex = re.compile(r'"(?:[^"]|"")*"')

This just finds the literals, it doesn't decode them by replacing the doubled quotes.

answered Jun 3, 2013 at 18:48

morningstar

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Comments

Jon Clements · Accepted Answer · 2013-06-03 18:48:12Z

Not using a regular expression, but you've specified Python, so here's a way to get your expected output:

>>> import csv >>> strings = ['"some string"', '"hello "" there"', '"hello """" there"'] >>> for s in strings: print next(csv.reader([s])) ['some string'] ['hello " there'] ['hello "" there']

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