获取 src 属性时出错.(关于如何正确获取路径的问题) [英] Error on getting src attribute. ( question on how to get the path correctly )

问题描述

我正在尝试从 RSS 提要获取一些数据，但在获取 img 的 src 属性时遇到问题.该图像位于 //item/description/div[@class="separator"]//img 并且我所做的一项尝试是加入具有

I am trying to get some data from a RSS feed but I am having problem on getting the src attribute of img. The image is located at //item/description/div[@class="separator"]//img and one attempt I made is to join the posts xPath having

推荐答案

那里:

$imagespath = $IMAGES->item(0)->nodeValue;

您正在将某些 标记的内容放入 $imagespath 变量中.
因此，该变量不包含 DOMElement 或任何类型的对象，而是一些字符串内容.

You are getting the content of some <description> tag into the $imagespath variable.
That variable, so, doesn't contain a DOMElement nor any kind of object, but some string content.

然后，您尝试对该变量调用 getAttribute() 方法:

$image = $imagespath->getAttribute('src'); 

但是由于 $imagespath 不是对象(它是一个字符串)，你会得到一个致命错误.

But as $imagespath is not an object (it's a string), you get a Fatal Error.

如果你想获得某个标签的 src 属性，我想你应该用这样的东西来获得它:

If you want to get the src attribute of some tag, I suppose you should get it with something like this :

$image = $IMAGES->item(0)->getAttribute('src'); 

即读取 XPath 查询返回的节点的 src 属性——当然，在调用 getAttribute() 之前，您应该检查查询是否确实返回了一个项目.

i.e. reading the src attribute of a node returned by your XPath query -- of course, before calling getAttribute(), you should check that there is actually an item returned by the query.

另外，如果你想加载图片，也许你应该得到 标签:

$IMAGES = $post->getElementsByTagName( "img" );

而不是 的?

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