FAQ,
I have a small question: How can we list distinct
nodes ( without redundancy)?
In the example below, i want for example to list
manufacturers (mercedes,peugeot,Renault)with no
redundancy.
<xsl:key name="x" match="car" use="@manufacturer"/>
<xsl:template match="cars">
<xsl:copy>
<xsl:apply-templates select="car[generate-id() = generate-id(key('x',
@manufacturer))]"/>
</xsl:copy>
</xsl:template>
<xsl:template match="@* | node()">
<xsl:copy>
<xsl:apply-templates select="@* | node()"/>
</xsl:copy>
</xsl:template>
See Jeni's XSLT Pages on grouping <http://www.jenitennison.com/xslt/grouping/>.
Cheers,
Jarno - E-Craft: Brich Es! (Lights of Euphoria Mix)
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list