The record holds a pageid, which is the id of the page the record was submitted on, you can then use the library method umbraco.library:NiceUrl to fetch the full url
It turned out that the url parameter is already in the @records xml, it's an attribute of the pageid element, as per the xml in my original post. This must have been added in at some point. So all I had to do was replace this line in my XSLT file:
I didn't need to access the umbraco.library at all. I assume it's safe to use the url attribute of the pageid element for this version of Contour. I'd appreciate it if you'd let me know otherwise.
Getting url of page submitted from in xslt email workflow
Contour 1.1.12
I can't see how to display the url from this xml:
Into this xslt file (see Url:...):
What am I missing / doing wrong? I can't find out what parameters contour passes into the XSLT, only examples online that contain $records
Comment author was deleted
To get an overview of the xml format check out this doc page http://our.umbraco.org/projects/umbraco-pro/contour/documentation/Developer/Working-with-Contour-data-in-XSLT/
The record holds a pageid, which is the id of the page the record was submitted on, you can then use the library method umbraco.library:NiceUrl to fetch the full url
Hi @Tim
It turned out that the url parameter is already in the @records xml, it's an attribute of the pageid element, as per the xml in my original post. This must have been added in at some point. So all I had to do was replace this line in my XSLT file:
With this:
I didn't need to access the umbraco.library at all. I assume it's safe to use the url attribute of the pageid element for this version of Contour. I'd appreciate it if you'd let me know otherwise.
Thanks!
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