Re: AW: recursion with xsl:apply-templates

it is like David assumed:
> I'm not sure quite what transform you wanted, but i think it's just an
> identity transform plus copying the  @index'th id element from the
> idlist??

Thanks,
Volker.

Dear Volker,

instead of <xsl:variable name="p" select="count(preceding::*[@index])+1"/>
do you mean this?
<xsl:variable name="p" select="@index"/>
<xsl:copy-of select="/transformation/id_list/id[$p]"/>
In your sample xml you have two rec... with the same value of @index.
Will they get the same id or not? Can you use the index?

Cheers,

Juergen

-----Ursprüngliche Nachricht-----
Von: Volker Witzel [mailto:volker.witzel@xxxxxxxxxxxx]
Gesendet am: Mittwoch, 27. August 2003 15:10
An: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Betreff: Re:  recursion with xsl:apply-templates

Dear David, Mike, Andrew,

thanks very much for help and quick responses. Seems that I stuck on my recursion too much. Based on your input I'm now using the follwing code, which works as expected:

   <xsl:template match="/transformation">
     <xsl:apply-templates/>
   </xsl:template>
   <xsl:template match="@*|node()">
     <xsl:copy>
       <xsl:apply-templates select="@*|node()"/>
     </xsl:copy>
   </xsl:template>
   <xsl:template match="*[@index]">
     <xsl:copy>
       <xsl:copy-of select="@*[not(name()='index')]"/>
       <xsl:variable name="p" select="count(preceding::*[@index])+1"/>
       <xsl:copy-of select="/transformation/id_list/id[$p]"/>
       <xsl:copy-of select="child::node()"/>
     </xsl:copy>
   </xsl:template>
   <xsl:template match="id_list"/>

Davis: next time I'll provide a better input!

Volker.


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Mit freundlichen Grüßen / With kind regards
Volker Witzel
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