regex for string with backslash for escape

Question

I'm trying to come up with a pattern for finding every text that is between double or single quotation marks in java source code. This is what I have:

"(.*?)"|’(.*?)’

Debuggex Demo

This works for almost every case I guess except one:

"text\"moretext\"evenmore"

Debuggex Demo

This could be used as a valid String definition, because the quotes are escaped. The pattern does not recognize the inner part more text.

Any ideas for a pattern that accounts for this case?

Answer 1

You can use this regex to match single or double quotes string ignoring all escaped quotes:

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

RegEx Demo

RegEx Breakup:

• (["']): Match single or double quote and capture it in group #1
• (: Start Capturing group #2
  • [^\\]*?: Match 0 or more of any characters that is not a \
  • (?:`: Start non-capturing group
    • \\: Match a \
    • .: Followed by any character that is escaped
    • [^\\]*?: Followed by 0 or more of any non-\ characters
  • )*: End non-capturing group. Match 0 or more of this non-capturing group
• ): End capturing group #2
• \1: Match closing single or double quote matches in group #1

Answer 2

That should work: "([^"\\]|\\.)*"|'([^'\\]|\\.)*' Regexr test.

Explanation:

1. " matches ".
2. [^"\\]|\\. negates match of \ & "(i.e. makes it to consume \") or continues match of \ and any character.
3. * continue match.
4. " matches "

Same for '.