Re: [xsl] Apply for-each-group to a node subset

[Tony Graham <Tony.Graham@xxxxxxxxxxxxxxxxxxxxxx>]

On Fri, Sep 21 2007 16:28:47 +0100, rocketraman@xxxxxxxxxxx wrote:
> Tony Graham wrote:
>> On Fri, Sep 21 2007 08:51:13 +0100, rocketraman@xxxxxxxxxxx wrote:
>> ...
>> I'm still trying to get my head around what you want to do with your
>> template, but the likely reasons that your simplification doesn't work
>> are:
>
> I should have provided the output I wanted (obviously ignoring
> templates applied to irrelevant nodes). Here it is (basically grouping
> all nodes between g1 and g2 into a hierarchical structure).
>
> <r>
>   <n1/>
>   ...
>   <n2/>
>
>   <group>
>     <g1/>
>     ...
>     <g2/>
>   </group>
>
>   <group>
>     <g1/>
>     ...
>     <g2/>
>   </group>
>
>   <n3/>
>   ...
>   <n4/>
> </r>
>
> I stated in my original email, but didn't emphasize, that there are an
> arbitrary number of nodes between g1 - g2, n1 - n2, and n3 - n4 (my
> example nodes were badly named).

It may be simpler if you do it all using ">>":

  <xsl:template match="r">
    <xsl:copy>
      <xsl:apply-templates select="*[../g1[1] >> .]"/>
      <xsl:for-each-group
        select="g1[1] |
                *[(. >> ../g1[1]) and (../g2[last()] >> .)] |
                g2[last()]"
        group-ending-with="g2">
        <group>
          <xsl:apply-templates select="current-group()"/>
        </group>
      </xsl:for-each-group>
      <xsl:apply-templates select="*[. >> ../g2[last()]]"/>
    </xsl:copy>
  </xsl:template>


(The logic might be clearer if the predicates used "<<" instead, except
that "<<" has to be written as "&lt;&lt;".)

Regards,


Tony Graham.
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