我怎样才能获得 <yt:accesscontrol> 的 xml 属性? [英] How can i get to xml attribute of <yt:accesscontrol>?

问题描述

我无法弄清楚或找出如何使用 PHP simplexml_load_file 从这个 xml 中解析 permission="allowed" 值.

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

基本结构是

<?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" />

如何获取最后一行 permission=allowed 属性的值?

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

推荐答案

这太烦人了……

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

我可以通过以下方式访问权限属性:

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

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

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