[xml] How can I find the link URL by link text with XPath?

I have a well formed XHTML page. I want to find the destination URL of a link when I have the text that is linked.

Example

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

I want an XPath expression such that if given programming questions site it will give http://stackoverflow.com and if I give it news it will give http://cnn.com.

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.

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.

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'".

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

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

if you are using html agility pack use getattributeValue:

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