Regex for quoted string with escaping quotes

Question

How do I get the substring " It's big \"problem " using a regular expression?

s = ' function(){  return " It\'s big \"problem  ";  }';     

Answer 1

A more extensive version of https://stackoverflow.com/a/10786066/1794894

/"([^"\\]{50,}(\\.[^"\\]*)*)"|\'[^\'\\]{50,}(\\.[^\'\\]*)*\'|“[^”\\]{50,}(\\.[^“\\]*)*”/   

This version also contains

1. Minimum quote length of 50
2. Extra type of quotes (open “ and close ”)

Answer 2

Messed around at regexpal and ended up with this regex: (Don't ask me how it works, I barely understand even tho I wrote it lol)

"(([^"\\]?(\\\\)?)|(\\")+)+"

Answer 3

If it is searched from the beginning, maybe this can work?

\"((\\\")|[^\\])*\"

Answer 4

As provided by ePharaoh, the answer is

/"([^"\\]*(\\.[^"\\]*)*)"/

To have the above apply to either single quoted or double quoted strings, use

/"([^"\\]*(\\.[^"\\]*)*)"|\'([^\'\\]*(\\.[^\'\\]*)*)\'/

Answer 5

"(?:\\"|.)*?"

Alternating the \" and the . passes over escaped quotes while the lazy quantifier *? ensures that you don't go past the end of the quoted string. Works with .NET Framework RE classes

Answer 6

This one works perfect on PCRE and does not fall with StackOverflow.

"(.*?[^\\])??((\\\\)+)?+"

Explanation:

1. Every quoted string starts with Char: " ;
2. It may contain any number of any characters: .*? {Lazy match}; ending with non escape character [^\\];
3. Statement (2) is Lazy(!) optional because string can be empty(""). So: (.*?[^\\])??
4. Finally, every quoted string ends with Char("), but it can be preceded with even number of escape sign pairs (\\\\)+; and it is Greedy(!) optional: ((\\\\)+)?+ {Greedy matching}, bacause string can be empty or without ending pairs!