Re: [xsl] Finding distinct nodes using a function in XSLT 2.0

[Abel Braaksma <abel.online@xxxxxxxxx>]

Dimitre Novatchev wrote:
> On 4/17/07, Scott Lynch <slynch@xxxxxxxxxx> wrote:
> > Dimitre,
> >
> > That works nicely.
> >
> > However, is there any difference between
> > <xsl:sequence select="."/>
> > (which works) rather than say
> > <xsl:sequence select="current-group()[1]"/>
> > which also works, in the overall template logic?
>
> Hi Scott,
>
> No there isn't any difference (in this case), because as the XSLT2.0 
> spec says:

But note that (the other case) you can use current-group() still when 
the context item has changed. I.e., in another template which is called 
or indirectly invoked from your for-each-group. In which case the dot 
'.' will contain something else then current-group()[1]

Btw, you can also do the distinct nodes in a single xpath (the title of 
your post seems to suggest you were looking for a function), as long as 
your identity is based on a value (of @id):

   <xsl:variable name="files"
         select="document(('file1.xml', 'file2.xml'))" />
     
   <xsl:template match="/" name="main">
       <xsl:copy-of select="
           for $i in distinct-values($files/*/*/@id)
           return ($files/*/*[@id = $i])[1]" />
   </xsl:template>


If you need sorting, you can use xsl:perform-sort, change the 
xsl:copy-of like this:

  <!-- or, with sorting -->
  <xsl:perform-sort select="
        for $i in distinct-values($files/*/bar/@id)
        return ($files/*/*[@id = $i])[1]" >

      <xsl:sort select="@id" />
  </xsl:perform-sort>


(btw: note the double parentheses around the fn:document() function, 
they are on purpose)

Cheers,
-- Abel Braaksma