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hi i am trying to get the extension of the file called in a url (eg /wp-includes/js/jquery/jquery.js?ver=1.3.2 HTTP/1.1) and get the query parameters passed to the file too.

What would be the best way to the extension?

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urlparse.urlparse() and os.path.splitext().

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You should probably just rsplit on . rather than use os.path in this case, as a URL and a filepath are different things. Using os.path.splitext would go wrong on an operating system where the file extension separator isn't .. Admittedly there are vanishingly few of those today.
wouldn't rsplit mathc the ver=1.3.2 path too ?
You would use it on the results of urlparse().
As a rule URL path manipulations should probably be done with posixpath.

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