I have an iFrame on my website that pops up and will play a video. The iframe source location is:
http://www.fivepointsusa.com/video.php?id=1&v=3_2.flv
The like button WILL pick this link up properly. However its a pretty bland page as it's designed to be inside a specified sized iframe.
The link that i WANT to show is:
http://www.fivepointsusa.com/bio.php?id=1#!prettyPhoto/0/
That link will bring the user to my bio page, open the iFrame and start the video. When I get my script to insert the correct address into the
However when i click the Like button, it drops off the #!prettyPhoto/0/ part. This is the most important part of the link...
How can i get the rest of开发者_开发百科 the URL to show?
You can't. Web browsers are not supposed to transfer the fragment (the part after the '#') when requesting a resource from a server, so that part gets dropped. You can get around this by putting the '!prettyPhoto/0/' part in a URL parameter, e.g. '?id=1&path=prettyPhoto/0/' (make sure to URL encode the parameters).
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