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(_dot_)Elovirta(_at_)nokia(_dot_)com
[mailto:Jarno(_dot_)Elovirta(_at_)nokia(_dot_)com]
Sent: Tuesday, 21 December 2004 8:42 PM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: RE: [xsl] 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
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