Any ideas how this function can return the linked URL when searching for an href using a wildcard?
I'm searching on a page for a backlink with an href of domain.com, but I'd like the tool to also provide me with what the actual href is that was linked to. Example: https://domain.com/service/page
Any ideas?
That's a great question. I'm wondering why we hadn't thought of this use case before, or perhaps I'm missing an easier method.
Here is an XPath that should get you the desired results:
Thank you for the reply!
This is close to what I'm wanting to do, but the below image is more what I'm going for. I'm not too familiar with the Xpath functions. Can you help me with the example shown below?
Yes, the formula will work for that purpose:
Basically, the XPathOnUrl function requests text from the HTML in the URL in column B. The "xpath" (//a...) is a syntax to specify what text to extract. If you are familiar with HTML then you know that a is a hyperlink and //a means selecting all links. The contains() filters the links to only those having an href attribute containing "tesla.com" (column A). Then finally the third argument "href" means to get the attribute, not the text. Here is a good link if you want to learn more about Xpath.
The Xpath function will return all matches and if you want to output all links, then you can wrap it around the Dump() function. I set the transpose argument to true which outputs the links horizontally instead of vertically. This is useful when requesting more than one value for every row:
Remove the @ sign in the beginning. I use that because I'm on the latest Office 365 version with "array formulas" that outputs all results. The @ sign makes only the first value appear.
