Subject: Re: [xsl] xsl-for-each-group in xslt 2.0 From: Andrew Welch <andrew.j.welch@xxxxxxxxx> Date: Mon, 28 Sep 2009 16:01:03 +0100 |
2009/9/28 a kusa <akusa8@xxxxxxxxx>: > Hi > > I am having difficulties implementing xsl-for-each-group in xslt 2.0 > > My XML is as follows: > > <list1> > <listitem>A</listitem> > <listitem>B</listitem> > <list2> > <listitem>C</listitem> > <listitem>D</listitem> > </list2> > </list1> > > Every <listX><listitem> combination corresponds to one step. > > So I want this to transform into: > > <step1>A B</step1> > <step2>C D</step2> > > I tried > > <xsl template match="list1"> > > <xsl:-for-each-group select="." group-by="listitem"> > <step1><xsl:apply-templates/></step1> > > </xsl:for-each-group> > <xsl:template> > > Can someone please help me understand where i am going wrong here? It's not actually a grouping problem, the following templates will give the output you want: <xsl:template match="list1"> <step1> <xsl:value-of select="listitem"/> </step1> <xsl:apply-templates select="list2"/> </xsl:template> <xsl:template match="list2"> <step2> <xsl:value-of select="listitem"/> </step2> <xsl:apply-templates select="list3"/> </xsl:template> If the nesting depth is unknown you can modify that to be a single template with name tests... -- Andrew Welch http://andrewjwelch.com Kernow: http://kernowforsaxon.sf.net/
Current Thread |
---|
|
<- Previous | Index | Next -> |
---|---|---|
[xsl] xsl-for-each-group in xslt 2., a kusa | Thread | Re: [xsl] xsl-for-each-group in xsl, Martin Honnen |
[xsl] xsl-for-each-group in xslt 2., a kusa | Date | Re: [xsl] xsl-for-each-group in xsl, Martin Honnen |
Month |