Matching Quote Wrapped Strings In Javascript With Regex

I need a regex for javascript for matching '{any group of chars}' <-- where that last ' is not preceeded by a \ examples: ... foo 'bar' ... => 'bar' ... foo'bar\'' ... =&g

Solution 1:

Prioritize the escaped characters first:

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

https://regex101.com/r/sj4HXw/2

Solution 2:

Simple scenario (as in OP)

The most efficient regex (that is written in accordance with the unroll-the-loop principle) you may use here is

"[^"\\]*(?:\\[\s\S][^"\\]*)*"

See the regex demo

Details:

• " - match the first "
• [^"\\]* - 0+ chars other than " and \
• (?:\\[\s\S][^"\\]*)* - zer or more occurrences of:
  • \\[\s\S] - any char ([\s\S]) with a \ in front
  • [^"\\]* - 0+ chars other than " and \
• " - a closing ".

Usage:

// MATCHINGvar rx = /"[^"\\]*(?:\\[\s\S][^"\\]*)*"/g;
var s = '    ... foo "bar" ...  goo"o"ooogle "t\\"e\\"st"[]';
var res = s.match(rx);
console.log(res);

// REPLACINGconsole.log(s.replace(rx, '<span>$&</span>'));

More advanced scenario

If there is an escaped " before a valid match or there are \s before a ", the approach above won't work. You will need to match those \s and capture the substring you need.

/(?:^|[^\\])(?:\\{2})*("[^"\\]*(?:\\[\s\S][^"\\]*)*")/g
 ^^^^^^^^^^^^^^^^^^^^^^                             ^

See another regex demo.

Usage:

// MATCHINGvar rx = /(?:^|[^\\])(?:\\{2})*("[^"\\]*(?:\\[\s\S][^"\\]*)*")/g;
var s = '    ... \\"foo "bar" ...  goo"o"ooogle "t\\"e\\"st"[]';
var m, res=[];
while (m = rx.exec(s)) {
  res.push(m[1]);
}
console.log(res);

// REPLACINGconsole.log(s.replace(/((?:^|[^\\])(?:\\{2})*)("[^"\\]*(?:\\[\s\S][^"\\]*)*")/g, '$1<span>$2</span>'));

The main pattern is wrapped with capturing parentheses, and this is added at the start:

• (?:^|[^\\]) - either start of string or any char but \
• (?:\\{2})* - 0+ occurrences of a double backslash.

Solution 3:

This should do it:

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

It's a mixture of the other answers from Wiktor and Taufik.