I want to take an image URL from an API. The API only contains the dynamic part of the URL, without the domain. I have used the following formula: “Domain” + [URL].
The example result is exactly what I was looking for. But somehow, the example result is not being recognized as an URL.
When I take the example result and put it in quotation marks as the new formula, the image loads as expected. How do I fix this?
Didn’t mean to edit this haha, but I was suggesting to use the URL() function.
Thanks for the quick reply. The URL() function does convert it to a “URL”. But that does not solve the problem, because I need a “web-url”.
Does it still work even though it’s giving you an error? I am able to see the image I’m testing with even though it says it’s not a valid web-url.
Not the perfect reply, but this could just be a bug.
Ah yes, it does work with the error. Thank you!