RE: [xsl] listing nodes without redundancy

Subject: RE: [xsl] listing nodes without redundancy
From: Jarno.Elovirta@xxxxxxxxx
Date: Mon, 20 Oct 2003 10:35:30 +0300

> 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:apply-templates select="car[generate-id() = generate-id(key('x', @manufacturer))]"/>
  <xsl:template match="@* | node()">
      <xsl:apply-templates select="@* | node()"/>

See Jeni's XSLT Pages on grouping <>.


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