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

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'".

Answer 2

if you are using html agility pack use getattributeValue:

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

Answer 3

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.

Answer 4

Should be something similar to:

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

Answer 5

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

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

Answer 6

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.