Regex for quoted string

Question

How to get the substring " It big \"problem " using regular expression?

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

Answer 1

 /"(?:[^"\\]|\\.)*"/ 

Works at Regex Coach and PCRE Workbench.

JavaScript test example:

  var s = ' function(){ return " Is big \\"problem\\", \\no? "; }'; var m = s.match(/"(?:[^"\\]|\\.)*"/); if (m != null) alert(m); 

Answer 2

This comes from nanorc.sample, available on many linux distributions. It is used to highlight the syntax of C-style strings.

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

Answer 3

As stated in ePharaoh, the answer is

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

To apply the above to single quotes or double quotes, use

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

Answer 4

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

Alternating \" and . Goes through escaped quotes, while the lazy *? Quantifier ensures you don't go past the end of the quoted string. Works with .NET Framework RE classes

Answer 5

Most proposed solutions use alternative repetition paths, i.e. (A | B) *.

You may encounter stack overflow on large inputs, as some template compiler implements this using recursion.

Java, for example: http://bugs.java.com/bugdatabase/view_bug.do?bug_id=6337993

Something like this: "(?:[^"\\]*(?:\\.)?)*" , Or the one provided by Guy Bedford will reduce the number of parsing steps, avoiding most stack overflows.

Answer 6

 /"(?:[^"\\]++|\\.)*+"/ 

Taken directly from man perlre on a Linux system with Perl 5.22.0 installed. As an optimization, this regular expression uses the "posessive" form of both + and * to prevent backtracking, since it is known in advance that a line without a closing quote will in no way match.

Answer 7

 /(["\']).*?(?<!\\)(\\\\)*\1/is 

should work with any quoted string

Answer 8

This works fine on PCRE and does not crash with StackOverflow.

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

Explanation:

• Each quoted string begins with Char: " ;
• It can contain any number of characters:. .*? {Lazy match}; ending with non escape [^\\] ;
• The operator (2) is Lazy (!) Optional, because the string may be empty (""). So: (.*?[^\\])??
• Finally, each quoted string ends with Char ( " ), but it may be preceded by an even number of escape code pairs (\\\\)+ ; and this Greedy (!) Is optional: ((\\\\)+)?+ {Greedy matching}, the bacause string may be empty or without trailing pairs!

Answer 9

here is what works with both “and” and easily adds others at the beginning.

  ("| ') (?: \\\ 1 | [^ \ 1]) *? \ 1 

it uses a backlink (\ 1) to exactly match what is in the first group ("or").

http://www.regular-expressions.info/backref.html

Answer 10

Keep in mind that regular expressions are not a silver bullet for all -y strings. Some things are easier to do with the cursor and linear, manual search. A CFL could do the trick quite trivially, but there are not many CFL implementations (afaik).

Answer 11

If the search is done from the very beginning, maybe this can work?

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

Answer 12

A more extensive version of https://stackoverflow.com/a/312618/

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

This version also contains

• Minimum Quote Length 50
• Additional types of quotes (open " and close " )

Answer 13

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

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