regex - ignore escaped chars within quotation marks

Question

I am working with regular expressions for years now but this time I am struggling hard.

Assuming I have the following simple string.

$content = "foo 'bar \' [\'],' test";

Now I want to match everything within single quotation marks. There are even escaped characters in it. With my current regex those escaped characters are blowing up the results.

$regex = "/\'[^\'\\\\]*(?:\\\\.[^\'\\\\]*)*\'/";

The results

// expected
[0] => "'bar \' [\'],'"

// current
[0] => "'],'"

Live demo HERE .

Answer 1

You can try like this one

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

Demo

Or even more precise

'([^']|(\'))*?(?<!\\)'

Demo

Answer 2

You want here to find an open quote and its closing one, so no escaped quote.
(?<!\\\\)'.*?(?<!\\\\)' will do so
Explanation :

(?<! negative lookbehind
\\\\) escaped backslash and closing lookbehind
' the quote which has not been escaped (the negative look behind has checked it)
.*? any char : .* in lazy mode : ? so the next quote will be evaluate
(?<!\\\\) again negative lookbehind to check if the quote has been escaped
' Final not escaped quote