我怎样才能获得<yt：accesscontrol>的xml属性？

Question

I can't figure out or find out how to parse the permission="allowed" value out of this xml, using PHP simplexml_load_file.

the basic structure is

<?xml version='1.0' encoding='UTF-8'?>
<entry xmlns="http://www.w3.org/2005/Atom" xmlns:media="http://search.yahoo.com/mrss/" xmlns:gd="http://schemas.google.com/g/2005" xmlns:yt="http://gdata.youtube.com/schemas/2007" gd:etag="W/&quot;DkEDSH47eCp7I2A9WhJbEEQ.&quot;">
<yt:accessControl action="comment" permission="allowed" />
<yt:accessControl action="commentVote" permission="allowed" />
<yt:accessControl action="videoRespond" permission="moderated" />
<yt:accessControl action="rate" permission="allowed" />
<yt:accessControl action="embed" permission="allowed" />
<yt:accessControl action="list" permission="allowed" />
<yt:accessControl action="autoPlay" permission="allowed" />
<yt:accessControl action="syndicate" permission="allowed" />

How can get to the value of the permission=allowed attribute on that last line?

Answer 1

This was so annoying to figure out...

(where $xml = simplexml_load_file($source); )

I'm able to get to the permission attribute with:

$xml->children('http://gdata.youtube.com/schemas/2007')->accessControl[4]->attributes()->permission;