我正在使用简单的代码阅读 rss 提要:
<?php
$homepage = file_get_contents('http://www.forbes.com/news/index.xml');
$movies = new SimpleXMLElement($homepage);
echo '<pre>';
print_r($movies);
?>
输出如下: SimpleXMLElement 对象 ( [@attributes] => 数组 ( [版本] => 2.0 )
[channel] => SimpleXMLElement Object
(
[title] => SimpleXMLElement Object
(
)
[link] => SimpleXMLElement Object
(
)
[description] => SimpleXMLElement Object
(
)
[language] => en-us
[copyright] => Copyright 2009 Forbes.com LLC
[item] => Array
(
[0] => SimpleXMLElement Object
(
[title] => SimpleXMLElement Object
(
)
[link] => SimpleXMLElement Object
(
)
[author] => SimpleXMLElement Object
(
)
[pubDate] => Sat, 05 Nov 2011 07:17:21 GMT
[description] => SimpleXMLElement Object
(
)
)
还有更多…… 但是当我查看此页面的源代码时,我有这样的信息:
<rss version="2.0"><channel><title><![CDATA[Forbes.com: News]]></title><link><! [CDATA[http://www.forbes.com]]></link><description><![CDATA[News and reports from Forbes.com]]></description><language>en-us</language><copyright>Copyright 2009 Forbes.com LLC</copyright><item><title><![CDATA[Benicio Del Toro Offered Villain Role In "Star Trek" Sequel - Is It Khan?]]></title><link><![CDATA[http://www.forbes.com/sites/markhughes/2011/11/05/benicio-del-toro-offered-villain-role-in-star-trek-sequel-is-it-khan/?feed=rss_home]]></link><author><![CDATA[Mark Hughes]]></author><pubDate>Sat, 05 Nov 2011 07:17:21 GMT</pubDate><description><![CDATA[Variety reports that actor Benicio del Toro is being offered the role of villain in the upcoming sequel to director J.J. Abram?s 2009 blockbuster franchise-reboot movie Star Trek. So far, Abrams and crew have kept a tight lid on details about the new Paramount film, and the identity of the main villain is a closely ...]]></description>
如何在 mydatabase 中读取和存储 CDATA 值。
最佳答案
告诉 SimpleXML 将 CDATA 转换为普通文本:
$homepage = 'http://www.forbes.com/news/index.xml';
$movies = simplexml_load_file($homepage, "SimpleXMLElement", LIBXML_NOCDATA);
这应该为您完成,使用 simplexml_load_file
而不是 file_get_contents
。
关于php - 从 rss 提要中读取 cdata,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/8020181/