How to get opposite of a regex to find all content outside of any kind of matching quotes (ignoring escaped quotes and the 1st arg of ".replace()"

Asked Dec 17 '22 at 11:12

Active Dec 17 '22 at 14:29

Viewed 33 times

I have a regex that works, as seen in https://stackoverflow.com/a/74576808/1432181.

\(\s*/.*?/\w*\h*,(*SKIP)(*F)|(?|(")[^"\\]*(?:\\.[^"\\]*)*"|(')[^'\\]*(?:\\.[^'\\]*)*')

This selects any kinds of content inside matching quotes, ignoring escaped quotes, ignoring when in the first argument of a .replace() [which may be things like .replace(/'/g, ...)

(See https://regex101.com/r/42rHQx/1)

But how can I get the opposite of this (i.e. where it's blue in the regex101, I want it not selected - and where it was previously not selected I want it blue)?

(I need to use this in a preg_replace so that I can only replace things outside of the shown selected [quoted] content)

edited Dec 17 '22 at 14:29

Wiktor Stribiżew

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asked Dec 17 '22 at 11:12

user1432181

One option could be replacing the match with an empty string. – The fourth bird Dec 17 '22 at 11:15
you can use the following regex to select everything that is not inside matching quotes, ignoring escaped quotes: (?!'.*?'|".*?").* – Ritik Banger Dec 17 '22 at 12:22
Why not `preg_split` with this pattern of yours? – Wiktor Stribiżew Dec 17 '22 at 14:29
.. I want to str_replace the content that is not "quoted"... split will only show me the "blue" parts of the subject. See https://onlinephp.io/c/234c7 – user1432181 Dec 17 '22 at 15:54
Try it like this `("[^"\\\/]*(?:\\.[^"\\\/]*)*"|'[^'\\\/]*(?:\\.[^'\\\/]*)*')(*SKIP)(*FAIL)|.+?(?=(?1)|$)` https://regex101.com/r/zbm5lB/1 – The fourth bird Dec 26 '22 at 10:52

How to get opposite of a regex to find all content outside of any kind of matching quotes (ignoring escaped quotes and the 1st arg of ".replace()"

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