一尘不染

如何使用SimpleXML获取具有名称空间的节点的属性?

php

youtube.xml

<feed xmlns="http://www.w3.org/2005/Atom" xmlns:media="http://search.yahoo.com/mrss/" xmlns:openSearch="http://a9.com/-/spec/opensearchrss/1.0/" xmlns:gd="http://schemas.google.com/g/2005" xmlns:yt="http://gdata.youtube.com/schemas/2007">

    <entry>
        ...
        <yt:duration seconds="1870"/>
        ...
    </entry>

</feed>

update_videos.php

$source = 'youtube.xml';

// load as file
$youtube = new SimpleXMLElement($source, null, true);

foreach($youtube->entry as $item){
    //title works
    echo $item->title;

    //now how to get seconds? My attempt...
    $namespaces = $item->getNameSpaces(true);
    $yt = $item->children($namespaces['yt']);
    $seconds = $yt->duration->attributes();
    echo $seconds['seconds'];
    //but doesn't work :(
}

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2020-05-29

共1个答案

一尘不染

所以我找到了一种使用xpath的方法,这是最好的方法还是与我的问题代码保持一致的方法?只是出于好奇。

$source = 'youtube.xml';

// load as file
$youtube = new SimpleXMLElement($source, null, true);
$youtube->registerXPathNamespace('yt', 'http://gdata.youtube.com/schemas/2007');

$count = 0;
foreach($youtube->entry as $item){

    //title works
    echo $item->title;

    $attributes = $item->xpath('//yt:duration/@seconds');
    echo $attributes[$count]['seconds'];
    $count++;
}
2020-05-29