Question

我正在尝试开发包含一些YouTube视频的网站。从API中检索XML文件后，我有以下内容。

    <?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'>
    <id>http://gdata.youtube.com/feeds/api/videos/4ZsiqqOyWx8</id>
    <published>2007-08-03T05:48:51.000Z</published>
    [...]
    <author>
        <name>ak326</name>
        <uri>http://gdata.youtube.com/feeds/api/users/ak326</uri>
    </author>
    <gd:comments>
        <gd:feedLink href='http://gdata.youtube.com/feeds/api/videos/4ZsiqqOyWx8/comments' countHint='0'/>
    </gd:comments>
    <media:group>

        [...]
        <yt:duration seconds='222'/>
    </media:group>
    <gd:rating average='5.0' max='5' min='1' numRaters='4' rel='http://schemas.google.com/g/2005#overall'/>
    <yt:statistics favoriteCount='8' viewCount='2674'/>
</entry>

我正在尝试使用PHP检索此视频的长度，但使用

echo $xml->media->yt

但它不起作用。我认为它与媒体上的psuedo类有关，但我不知道如何选择它们。

Answer 1

这些XML元素是命名空间。您需要get the namespace information。

实施例

// get nodes in media: namespace for media information
$media = $entry->children('http://search.yahoo.com/mrss/');

// get video player URL
$attrs = $media->group->player->attributes();

Answer 2

尝试DOMXPath

$xml = new DOMDocument();
$xml->load(path/to/file);
$xpath = new DOMXPath($xml);
$xpath->registerNamespace("atom", "http://www.w3.org/2005/Atom");
$xpath->registerNamespace("media", "http://search.yahoo.com/mrss/");
$xpath->registerNamespace("yt", "http://gdata.youtube.com/schemas/2007");
print $xpath->query("/atom:entry/media:group/yt:duration/@seconds")->item(0)->value;

Answer 3

我假设你在这里使用SimpleXML

$nsMedia = $xml->children('http://search.yahoo.com/mrss/');
$group = $nsMedia->group;
$nsYt = $group->children('http://gdata.youtube.com/schemas/2007');
$duration = $nsYt->duration;
echo $duration['seconds'];

在PHP中检索XML伪类值？

3 个答案:

实施例