如何使用Xpath获取RSS URL(How do I pick up RSS URLs using Xpath)

如何使用Xpath获取RSS URL(How do I pick up RSS URLs using Xpath)

示例代码：

<link rel="alternate" type="application/rss+xml" title="Fit Living Lifestyle &raquo; Feed" href="http://www.fitlivinglifestyle.com/feed/" />

我如何检查type="application/rss+xml" XPATH，如果它存在，则返回URL： http://www.fitlivinglifestyle.com/comments/feed/ ： http://www.fitlivinglifestyle.com/comments/feed/ （甚至标题属性也是如此）

我尝试了一些变化：

//@href[parent::link[contains(@type='application/rss+xml',//@href)]]

但似乎并不完全存在。

Example code:

<link rel="alternate" type="application/rss+xml" title="Fit Living Lifestyle &raquo; Feed" href="http://www.fitlivinglifestyle.com/feed/" />

How do I have XPATH check for type="application/rss+xml" and if it is present then return me the URL: http://www.fitlivinglifestyle.com/comments/feed/ (and even the title attribute too)

I have tried some variations of this:

//@href[parent::link[contains(@type='application/rss+xml',//@href)]]

But don't seem to be quite there.

最满意答案

您正在寻找的XPath是：

link[@type="application/rss+xml"]/@href

这将选择type="application/rss+xml"所有link元素并返回href属性。 要返回title属性，请使用：

link[@type="application/rss+xml"]/@title

The XPath you are looking for is:

link[@type="application/rss+xml"]/@href

This selects all link elements with type="application/rss+xml" and returns the href attribute. To return the title attribute use:

link[@type="application/rss+xml"]/@title