如何使用 XPath 通过链接文本查找链接 URL？

我有一个良好的形式 XHTML页。我想找到一个链接的目的地 URL 时，我有文本被链接。

例子

<a href="http://stackoverflow.com">programming questions site</a>
<a href="http://cnn.com">news</a>

我想要一个 XPath表达式，如果给它 programming questions site，它将给 http://stackoverflow.com，如果我给它 news，它将给 http://cnn.com。

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最佳答案

Should be something similar to:

//a[text()='text_i_want_to_find']/@href

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//a[text()='programming quesions site']/@href

which basically identifies an anchor node <a> that has the text you want, and extracts the href attribute.

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Think of the phrase in the square brackets as a WHERE clause in SQL.

So this query says, "select the "href" attribute (@) of an "a" tag that appears anywhere (//), but only where (the bracketed phrase) the textual contents of the "a" tag is equal to 'programming questions site'".

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Too late for you, but for anyone else with the same question...

//a[contains(text(), 'programming')]/@href

Of course, 'programming' can be any text fragment.

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if you are using html agility pack use getattributeValue:

$doc2.DocumentNode.SelectNodes("//div[@class='className']/div[@class='InternalClass']/a[@class='InternalClass']").GetAttributeValue("href","")

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For case insensitive contains, use the following:

//a[contains(translate(text(),'PROGRAMMING','programming'), 'programming')]/@href

translate converts capital letters in PROGRAMMING to lower case programming.