使用PHP从XML获取标记值

My XML file looks like

我的XML文件看起来像

<entry>
    some tags 
    <title type='text'> content to be extracted </title>
    <content type='text'> content to be extracted </content>
    <link rel='alternate' type='text/html' href='https://www.youtube.com/watch?v=content to be extracted &amp;feature=youtube_gdata'/>
</entry>

to achieve this if I use the following ...

如果我使用以下内容来实现这一点......

$dom = new DOMDocument();
libxml_use_internal_errors(true);
$dom->loadXML("kFeed.xml");
$entryTag = $dom->getElementsByTagName("entry")->item(0)->nodeValue;
$title = $dom->$entryTag->getElementsByTagName("title")->item(0)->nodeValue;
$descr = $dom->$entryTag->getElementsByTagName("content")->item(0)->nodeValue;

It is throwing the following error like Notice: Trying to get property of non-object for 4th line and Fatal error: Cannot access empty property for 5th line.

它抛出以下错误，如注意：尝试获取第4行的非对象属性和致命错误：无法访问第5行的空属性。

Can someone please suggest me how to crack this and how to extract the 'v' value from the href of link (I have 4 links and I want that value which has rel='alternate').

有人可以建议我如何破解这个以及如何从链接的href中提取'v'值（我有4个链接，我想要那个具有rel ='alternate'的值）。

And finally when I directly give an URL (instead kFeed.xml) to load XML it is not functioning, no error.

最后，当我直接给出一个URL（而不是kFeed.xml）来加载XML时，它没有运行，没有错误。

4 个解决方案

#1

loadxml($xml) expects $xml to be the xml data itself, not a pointer to a (file) source.
Use DOMDocument::load($filename) instead.

loadxml（$ xml）期望$ xml是xml数据本身，而不是指向（文件）源的指针。请改用DOMDocument :: load（$ filename）。

And $dom->$entryTag->... won't work either. You might be interested in DOMXPath or SimpleXML instead.

并且$ dom - > $ entryTag - > ...也不起作用。您可能对DOMXPath或SimpleXML感兴趣。

#2

This will work for you

这对你有用

$str    =   "<entry>
some tags 
<title type='text'> content to be extracted </title>
<content type='text'> content to be extracted </content>
<link rel='alternate' type='text/html' href='https://www.youtube.com/watch?v=content to be extracted &amp;feature=youtube_gdata'/>
</entry>";

$xml = simplexml_load_string($str);

echo $xml->title;    //Will print title

echo "<pre>";
print_r($xml->link->attributes());   //Will print attributes also

#3

Should you actually be accessing the item as an array rather than as an object?

你真的应该作为一个数组而不是一个对象访问该项目吗？

$descr = $dom->$entryTag->getElementsByTagName("content")->item[0]->nodeValue;

#4

$xml = simplexml_load_file('xx.xml');

$title = (string)$xml->title;
$content = (string)$xml->content;
$linkHref = (string)$xml->link['href'];

$url = parse_url($linkHref);
parse_str($url['query'], $params);

echo '<pre>'; print_r($params['v']); echo '</pre>';

#1