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GREP to find a sequence of characters without a closing apostrophe

I'm trying to find long quotes in the text that I'm editing so that I can apply a different style to them. I've tried this GREP:

~[.{230}(?!.~])

What I need is for the GREP to find any 230 characters preceded by a left/opening quote mark, not including any 230-character sequence including a character followed by a right/clsoing quote mark. This should then eliminate quotes of less than 230 characters from the search. My GREP finds the correct length sequence but doesn't exclude those sequences which include a right quote mark.

So I want to find this, which my GREP does:

enter image description here

But not this, which my GREP also finds:

enter image description here

Because it has a closing quote in it and is therefore what I'm classing as a short quote.

Any ideas? TIA

Solution

  • You can match an opening ‘ followed by 230 or more occurrences of any character except an opening or closing quotation mark.

    To not match the closing quotation mark, you can assert it using a positive lookahead.

    ‘[^‘’]{230,}(?=’)
    • ‘ Match ‘
    • [^‘’]{230,} Repeat 230+ times any char except ‘ or ’ using a negated character class
    • (?=’) Positive lookahead, assert ’ directly to the right

    See a regex demo.