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Re: XSLT 2 and XSLT 3: Best Way To Get Set of Unique D

Subject: Re: XSLT 2 and XSLT 3: Best Way To Get Set of Unique Docs for A Set of Elements
From: "Graydon graydon@xxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>
Date: Tue, 5 Mar 2019 19:20:51 -0000
Re:  XSLT 2 and XSLT 3: Best Way To Get Set of Unique D
On Tue, Mar 05, 2019 at 07:08:02PM -0000, Eliot Kimber ekimber@xxxxxxxxxxxx scripsit:
> Given the variable $links that is a sequence of element()s where the elements could be from different documents, what is the best way to get the set of unique documents?
> 
> I need an XSLT 2 answer but an XSLT 3 answer would also be useful.
> 
> I feel like there's an obvious solution I'm overlooking but the only thing I can think of is to compare the document-uri() values of all the elements:
> 
>     <xsl:variable name="topicURIs" as="xs:string*"
>       select="distinct-values(for $e in $links return document-uri(root($e)))"
>     /> 
>     <xsl:variable name="topics" as="document-node()*"
>         select="
>         for $uri in $topicURIs
>         return root(($links[document-uri(root(.)) eq $uri])[1])
>         "
>     />
> 
> This works but seems unnecessarily complicated.

I will admit I'm not sure I follow just what you want, but how would

<xsl:variable name="uniqueTopics" as="document-node()*">
    <xsl:for-each-group select="$links" group-by="document-uri(root(.))">
       <xsl:sequence select="doc(current-grouping-key())" />
    </xsl:for-each-group>
</xsl:variable>

do?

-- Graydon

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