简单的 xpath 问题让我发疯 [英] Simple xpath question that drives me crazy

问题描述

以下是我设法使用此 xpath 打印内容的提要结构
$xml->xpath('/rss/channel//item')

结构

<;/link><author></author></item></channel></rss>

但是我的一些文件遵循这种结构

<feed xmlns="http://www.w3.org/2005/Atom" .....><entry><published></published><title></title><description></description><link></link><author></author></entry></feed>

我猜这应该是获取条目内容的xpath

$xml->xpath('/feed//entry')

证明我错的东西.

我的问题是什么是正确的 xpath 使用?我还缺什么吗?

这是代码

xpath('/feed//entry'));}echo "
";打印_r($条目)；回声"</pre>";?>
解决方案
试试这个:
$xml->registerXPathNamespace('f', 'http://www.w3.org/2005/Atom');$xml->xpath('/f:feed/f:entry');
below is the structure of a feed I managed to print the content using this xpath

$xml->xpath('/rss/channel//item')

the structure
<rss><channel><item><pubDate></pubDate><title></title><description></description><link></link><author></author></item></channel></rss>
However some of my files follow this structure
<feed xmlns="http://www.w3.org/2005/Atom" .....><entry><published></published><title></title><description></description><link></link><author></author></entry></feed>
and I guessed that this should be the xpath to get the content of entry
$xml->xpath('/feed//entry')
something that proved me wrong.

My question is what is the right xpath to use? Am i missing something else ?

This is the code
<?php

$feeds = array('http://feeds.feedburner.com/blogspot/wSuKU');


$entries = array();
foreach ($feeds as $feed) {
    $xml = simplexml_load_file($feed);
    $entries = array_merge($entries, $xml->xpath('/feed//entry'));
}

echo "<pre>"; print_r($entries); echo"</pre>";

?>

解决方案
try this:
$xml->registerXPathNamespace('f', 'http://www.w3.org/2005/Atom');
$xml->xpath('/f:feed/f:entry');


                        
这篇关于简单的 xpath 问题让我发疯的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！