Is there any PHP function that defines whether a given link is a video, picture or neither.
For example:
$link = www.example.com/342kddd23
$a = phpfunction($link);
If $link is a video link; $a = 1, if $link is a picture link $a = 2, if $link is not defined (it could be a normal link) $a = 0.
Is that possible? If so, are there any existing functions like that?
分享 I don't believe such a library exists (there's no built-in function I'm aware of).
You could of course attempt to develop your own, but you'll endlessly be chasing your tail if you plan to go beyond simply checking to see if there's a known file extension such as ".mp4", ".png", etc. due to the fact that you'll need to construct an increasingly large list of matching URL schemas. (This would also imply that a given link may only return a single type of content, which is often not the case.)
Assuming that you're looking for video and image files themselves (and not simply URLs that contain a video, etc.) an alternative (more accurate) approach would be request each link (perhaps using cURL with CURLOPT_HEADER enabled), see what content headers are returned (specifically "Content-Type") and take things from there.
Update
As a follow-up based on @smassey's excellent comment, setting the CURLOPT_NOBODY option (once again via curl_setopt) will send a HEAD request, hence ensuring you don't actually download any of the content beyond the response headers themselves.
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