Regex to match 0 or an even number of consecutive characters?

Question

I'm looking to see if it's possible to have a regex pattern that matches content between enclosing single quotes in a string, with the requirement that single quotes are escaped using another single quote.

Example:

The quick 'brown fox ''jumped ''''over the lazy dog''';

The regex should capture this string: 'brown fox ''jumped ''''over the lazy dog'''. Since a single quote is escaped using another single quote here, the rest of the string isn't included.

This is what I have so far (?<!\')\'(?!\').+(?<!\')\'(?!\')

This almost works, except the group isn't captured if the closing non-escaped single quote has an escaped quotes before it.

Is it possible to change the negative lookbehind to say that a single quote should be matched if there are either 0 or an EVEN number of single quotes behind it?

Answer 1

You may use this regex without any look around:

'((?:[^']+|'')*)'

RegEx Demo

RegEx Breakup:

• ': Match a '
• (: Start a capture group
  • (?::
    • [^']+: Match 1+ of any char that is not a '
    • |: OR
    • '': Match a pair of quotes i.e. ''
  • )*: End non-capture group. Repeat this group 0 or more times
• ): End capture group
• ': Match a '

Answer 2

I think it can also be unrolled:

'[^']*(?:''[^']*)*'

See this demo at regex101

• [^']* matches any amount of characters that are not a single quote
• (?: non capturing group ) for repeating what's inside