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@xxxxxxxxxxxx
<xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> 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.
>
> Thanks,
>
> Eliot
> --
> Eliot Kimber
> http://contrext.com
>
>



--
Cheers,
Dimitre Novatchev
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Truly great madness cannot be achieved without significant intelligence.
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To invent, you need a good imagination and a pile of junk
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Never fight an inanimate object
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To avoid situations in which you might make mistakes may be the
biggest mistake of all
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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
you're doing is work or play
-------------------------------------
To achieve the impossible dream, try going to sleep.
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Facts do not cease to exist because they are ignored.
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Typing monkeys will write all Shakespeare's works in 200yrs.Will they
write all patents, too? :)
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Sanity is madness put to good use.
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I finally figured out the only reason to be alive is to enjoy it.