Subject: Re: [xsl] XSLT 2 and XSLT 3: Best Way To Get Set of Unique Docs for A Set of Elements From: "Eliot Kimber ekimber@xxxxxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> Date: Tue, 5 Mar 2019 19:36:47 -0000 |
Of course--that's the obvious solution I was missing. Thanks, Eliot -- Eliot Kimber http://contrext.com o;?On 3/5/19, 1:15 PM, "Martin Honnen martin.honnen@xxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote: On 05.03.2019 20:08, Eliot Kimber ekimber@xxxxxxxxxxxx wrote: > 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]) > " > /> o;? > > This works but seems unnecessarily complicated. Wouldn't $links/root() suffice?
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Re: [xsl] XSLT 2 and XSLT 3: Best W, Martin Honnen martin | Thread | Re: [xsl] XSLT 2 and XSLT 3: Best W, Graydon graydon@xxxx |
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