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

Question

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

Answer 1

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.