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