Re: [xsl] xsl-for-each-group in xslt 2.0

[a kusa <akusa8@xxxxxxxxx>]

Hi

Thanks for all your responses.

Now I am able to get the listitems seperately under each step level.



On Mon, Sep 28, 2009 at 10:03 AM, Martin Honnen <Martin.Honnen@xxxxxx> wrote:
> a kusa wrote:
>
>> 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 am not sure you need xsl:for-each-group, for the above input (wrapped in
a
> root element) the following stylesheet creates above output (wrapped in a
> root element):
>
> <xsl:stylesheet
>  xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
>  version="2.0">
>
>  <xsl:output indent="yes"/>
>
>  <xsl:template match="root">
>    <xsl:copy>
>      <xsl:apply-templates/>
>    </xsl:copy>
>  </xsl:template>
>
>  <xsl:template match="*[matches(local-name(), 'list[0-9]+')]">
>    <xsl:element name="step{substring-after(local-name(), 'list')}">
>      <xsl:value-of select="listitem"/>
>    </xsl:element>
>    <xsl:apply-templates/>
>  </xsl:template>
>
>  <xsl:template match="text()"/>
>
> </xsl:stylesheet>
>
>
>
> --
>
>        Martin Honnen
>        http://msmvps.com/blogs/martin_honnen/