Subject: Re: [xsl] 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 |
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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