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I have the following question:

I'm modifying the XSLT of the search, so it is represented differently etc etc .. in this occasion I also need to get a relative site collection url.

In managed property i found a property SPSiteUrl, but adding it to properties for the xslt didn't help, as i got an error "inconsistent use of schema" etc etc, so how with other words can I retrieve the site collection relative url?

Because else i cannot know the relative url anyhow ...

OR if i'm in search, and i have the direct url to an item, how can i then by javascript decide which is the part of site collection url?

Any suggestions?

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1 Answer

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When I customize XSLT for list views and forms (either as XSL referenced in the List's schema, or as an *.ascx that I deploy to CONTROLTEMPLATES) I used the following pre-defined parameter. You just declare it.. and then use it.. it's actual value is supplied magically elsewhere (in main.xsl, I think)

 <xsl:param name="HttpVDir"/>

The above param declaration goes near the top of your stylesheet (between the xsl:stylesheet tags, usually preceding any specific xsl:template tags).

Then, if you put this within the context of an xsl:template, it will output the site collection URL:

 <xsl:value-of select="$HttpVDir" />
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Microsoft's official list of XSLT Global Parameters can be found here: msdn.microsoft.com/en-us/library/ff806158(v=office.14).aspx – bkwdesign Jan 3 at 22:42
So, to further answer your question, I would formulate the XSLT to output a tiny javascript block that declares a single JS variable: var siteCollectionUrl = ... which you can then reference in your other javascript logic. – bkwdesign Jan 3 at 22:46
In my further experimentation just now, for XSL files that are referenced in list schema documents, you can dispense with the param declaration and skip right to using the second snippet above. I think this is because XSL files start with an include reference to main.xsl – bkwdesign Jan 3 at 23:08

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