Question

我想从以下xml中选择标题和youtube链接：

`<?xml version="1.0" encoding="UTF-8"?><feed     xmlns="http://www.w3.org/2005/Atom"><category term="videos" label="/r/videos"/>    <icon>https://www.redditstatic.com/icon.png/</icon><id>/r/videos/.xml</id><link     rel="self" href="https://www.reddit.com/r/videos/.xml"     type="application/atom+xml" /><link rel="alternate" href="https://www.reddit.com/r/videos/" type="text/html" /><logo>https://a.thumbs.redditmedia.com/mtwnduVr0DnrK1o8rpTPi6waLWuPimj_8ntK8i5t890.png</logo><subtitle>A great place for video content of all kinds.</subtitle><title>Videos</title><entry><author><name>/u/LegendaryContent</name><uri>https://www.reddit.com/user/LegendaryContent</uri></author><category term="videos" label="/r/videos"/><content type="html">&lt;table&gt; &lt;tr&gt;&lt;td&gt; &lt;a href=&quot;https://www.reddit.com/r/videos/comments/45crp7/1400_employees_being_laid_off/&quot;&gt; &lt;img src=&quot;https://b.thumbs.redditmedia.com/UR4XFRqoMtj5watvSUrUlEdTYiA1gOv_OxqxtxNyftQ.jpg&quot; alt=&quot;1,400 Employees being laid off&quot; title=&quot;1,400 Employees being laid off&quot; /&gt; &lt;/a&gt; &lt;/td&gt;&lt;td&gt; &amp;#32; submitted by &amp;#32; &lt;a href=&quot;https://www.reddit.com/user/LegendaryContent&quot;&gt; /u/LegendaryContent &lt;/a&gt; &lt;br/&gt; &lt;span&gt;&lt;a href=&quot;https://youtu.be/Y3ttxGMQOrY&quot;&gt;[link]&lt;/a&gt;&lt;/span&gt; &amp;#32; &lt;span&gt;&lt;a href=&quot;https://www.reddit.com/r/videos/comments/45crp7/1400_employees_being_laid_off/&quot;&gt;[comments]&lt;/a&gt;&lt;/span&gt; &lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;</content><id>t3_45crp7</id><link href="https://www.reddit.com/r/videos/comments/45crp7/1400_employees_being_laid_off/" /><updated>2016-02-12T03:22:38+00:00</updated><title>1,400 Employees being laid off</title></entry></feed>`

我的代码在这里：

<?php
$videos ="";
$video_category = "Trending Videos";
$url = "https://www.reddit.com/r/videos/.xml";
$feed_dom = new domDocument; 
$feed_dom->load($url); 
$feed_dom->preserveWhiteSpace = false;
$items = $feed_dom->getElementsByTagName('entry');

foreach($items as $item){
$title = $item->getElementsByTagName('title')->item(0)->nodeValue;
$desc_table = $item->getElementsByTagName('content')->item(0)->nodeValue;

$table_dom = new domDocument;
$table_dom->loadHTML($desc_table);
$xpath = new DOMXpath($table_dom);
$table_dom->preserveWhiteSpace = false;
$yt_link_node = $xpath->query("//table/tr/td[2]/a[2]");

foreach($yt_link_node as $yt_link){

$yt = $yt_link->getAttribute('href');
echo $title;
echo $yt;
}
?>

出于某种原因，它无法正常工作，我已经应用了几乎所有在google＆amp; amp;堆栈溢出。标题很好，但不是$yt。你能选择我做错了吗？

Answer 1

这是因为DOM与您期望的略有不同。

您正在解析的HTML（ $ desc_table ）通常具有以下结构：

<table>
    <tr>
        <td>
            <a href="https://www.reddit.com/r/videos/comments/...">
                <img src="https://b.thumbs.redditmedia.com/....jpg" 
                     alt="..." title="..." />
            </a>
        </td>
        <td> &#32; submitted by &#32; 
            <a href="https://www.reddit.com/user/..."> /u/... </a> 
            <br/> 
            <span>
                <a href="https://youtu.be/...">[link]</a>
            </span>
            &#32;
            <span>
                <a href="https://www.reddit.com/r/videos/comments/.../">[comments]</a>
            </span> 
        </td>
    </tr>
</table>

因此没有第二个锚元素（a）是第二个td元素的直接子元素，因为第二个（和第三个）锚点包含在span标记中。

因此，如果您想要访问此链接：

                <a href="https://youtu.be/...">[link]</a>

然后使用此XPath：

 $yt_link_node = $xpath->query("//table/tr/td[2]/span[1]/a");

为什么XPath查询不起作用？

1 个答案: