Dear Volker,
instead of
<xsl:variable name="p" select="count(preceding::*[(_at_)index])+1"/>
do you mean this?
<xsl:variable name="p" select="@index"/>
<xsl:copy-of select="/transformation/id_list/id[$p]"/>
In your sample xml you have two rec... with the same value of @index.
Will they get the same id or not? Can you use the index?
Cheers,
Juergen
-----Ursprüngliche Nachricht-----
Von: Volker Witzel [mailto:volker(_dot_)witzel(_at_)schenker(_dot_)com]
Gesendet am: Mittwoch, 27. August 2003 15:10
An: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Betreff: Re: [xsl] recursion with xsl:apply-templates
Dear David, Mike, Andrew,
thanks very much for help and quick responses. Seems that I stuck on my
recursion too much. Based on your input I'm now using the follwing code,
which works as expected:
<xsl:template match="/transformation">
<xsl:apply-templates/>
</xsl:template>
<xsl:template match="@*|node()">
<xsl:copy>
<xsl:apply-templates select="@*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="*[(_at_)index]">
<xsl:copy>
<xsl:copy-of select="@*[not(name()='index')]"/>
<xsl:variable name="p" select="count(preceding::*[(_at_)index])+1"/>
<xsl:copy-of select="/transformation/id_list/id[$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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