Jarno, 

Thankyou very much for your help, I even managed to group to another level
now I've got my head around the concept.

Cheers,

Chris

> -----Original Message-----
> From: Jarno.Elovirta@xxxxxxxxx [mailto:Jarno.Elovirta@xxxxxxxxx] 
> Sent: Tuesday, 21 December 2004 8:42 PM
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Subject: RE:  Summarising XML datasets
> 
> Hi,
> 
> > I've tried following the example from Jeni's site and come 
> up with the
> > following:
> 
> snip
> 
> > <xsl:template match="results">
> >     <xsl:for-each select="photograph[count(. | key('idkey', 
> > id)[1])=1]">
> >         <h3><xsl:value-of select="id" /></h3>
> >         <h4><xsl:value-of select="name"/></h4>
> >         <h5><xsl:value-of select="description"/></h5>
> > 	<xsl:for-each select="key('subjectkey', id)">
> >             <TABLE border="0" width="75%">
> >                 <tr>
> >                     <th width="10%" align="right">Subject</th>
> >                     <td width="90%" align="left"><xsl:value-of 
> > select="subject" /></td>
> >                 </tr>
> >             </TABLE>
> >             <hr width="75%" align="left"/>
> >          </xsl:for-each>
> >   
> >     </xsl:for-each>
> > </xsl:template>
> > </xsl:stylesheet>
> > 
> > This works unless I have a duplicate subject node, which 
> actually can 
> > happen.
> > 
> > I'm a little bit stuck with how to only display unique subjects.
> 
> Then you need to generate a second grouping key, which is a 
> concatenation of ID and subject.
> 
>   <xsl:key name="subjectkey" match="photograph" 
> use="concat(id, ' ', subject)"/>
> 
> and list subjects with
> 
>   <xsl:for-each select="key('idkey', id)[generate-id(.) = 
> generate-id(key('subjectkey', concat(id, ' ', subject)))]">
>     <xsl:if test="not(position() = 1)">, </xsl:if>
>     <xsl:value-of select="subject" />
>   </xsl:for-each>
> 
> Cheers,
> 
> Jarno