Question

I need to match a string that is in quotations, but make sure the first quotation is not escaped.

For example: First \”string\” is “Hello \”World\”!”

Should match only Hello \”World\”!

I am trying to modify (")(?:(?=(\\?))\2.)*?"

I tried adding [^\\"] to ("), and that kinda works, but it matches either only (“) or every other letter that isn’t (\”) and I can’t figure out a way to modify ([\\"]") to only match (“) if it is not (\”)

This is what I have so far ([^\\"]")(?:(?=(\\?))\2.)*?"

I’ve been trying to figure it out using these two pages, but still cannot get it.

Can Regex be used for this particular string manipulation?

RegEx: Grabbing values between quotation marks

Thanks

Answer 1

You can use negative look behind like this:

(?<!\\)"(.*?)(?<!\\)"

Check see it in action here on regex101

The first match group contains:

Hello \"World\"!