RE: [xsl] listing nodes without redundancy

[Jarno.Elovirta@xxxxxxxxx]

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