To find the different document nodes just use:
$links/root(.)
Remember that in a node-set all nodes have different identity (this is
exactly what "set" means)
Cheers,
Dimitre
On Tue, Mar 5, 2019 at 11:08 AM Eliot Kimber ekimber(_at_)contrext(_dot_)com
<xsl-list-service(_at_)lists(_dot_)mulberrytech(_dot_)com> 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])
"
/>
This works but seems unnecessarily complicated.
Thanks,
Eliot
--
Eliot Kimber
http://contrext.com
--
Cheers,
Dimitre Novatchev
---------------------------------------
Truly great madness cannot be achieved without significant intelligence.
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To avoid situations in which you might make mistakes may be the
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Quality means doing it right when no one is looking.
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You've achieved success in your field when you don't know whether what
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