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Related question: How can I use regex to match a character (') when not following a specific character (?)?

I'm parsing a log using regex (PHP PCRE library), and trying to extract a URL from it. The URL is encapsulated in double quotes ", but some of the requests also include a double quote ". For example:

"https://www.amh.net.au/online/dbSearch.php?t=all&q=\"Rosuvastatin\""

My first pattern was basically:

#\"([^\"]*)\"#

This worked well, until I reached one of the entries as above, and it truncated the match so all I got was:

https://www.amh.net.au/online/dbSearch.php?t=all&q=\

After digging around, and rediscovering the cheatsheets for regex at http://addedbytes.com and also some more useful information at http://www.regular-expressions.info/lookaround.html I have now tried the following look-behind:

#"([(?<!\\)"]*)"#

But, now all I get is "" and then an empty string

Community
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HorusKol
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2 Answers2

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You placed your lookbehind INSIDE your group ([]), so it's not interpreted as such, but rather just you say you only want those individual characters.
Basically, I think you'd like something like this: 

#"(?:[^"]|(?<=\\)")"#

Though you should be aware that you'd be trolled by \\" for example.

Loamhoof
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The URLs in the logs would be URL-encoded. As such, the following pattern should work:

#\"([^ ]*)\"#
devnull
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