RE: [xsl] combining the children of similar nodes under one node

["Andreas L. Delmelle" <a_l.delmelle@xxxxxxxxxx>]

> -----Original Message-----
> From: David Montalvo
> 
<snip />
> One of the problems is that <style> and <type> are being used 
> interchangeably, but I think I compensated for that. My big problem is 
> getting everything listed under one name with no duplicates.
> 

Hi,

Sounds like a grouping-problem...

Try something like:

<xsl:stylesheet ...>
<xsl:key name="prods-by-name" match="product" use="name" />
<xsl:key name="style-type-by-prod" match="style | type"
         use="concat(parent::product/name,.)" />

<xsl:variable name="distinct-prods"
  select="//product[generate-id(.)=generate-id(
    key('prods-by-name',name))]" />

<xsl:template match="root">
  <xsl:apply-templates select="$distinct-prods" mode="merge"/>
</xsl:template>

<xsl:template match="product" mode="merge">
  <xsl:copy>
    <xsl:copy-of select="name" />
    <xsl:apply-templates select="key('prods-by-name',name)" mode="list" />
  </xsl:copy>
</xsl:template>

<xsl:template match="product" mode="list">
  <xsl:apply-templates select="(style | type)[generate-id(.)=generate-id(
                         key('style-type-by-prod',concat(
                           current()/name,.)))]" />
</xsl:template>

<xsl:template match="style | type">
  <xsl:copy-of select="." />
</xsl:template>

</xsl:stylesheet>


Hope this helps!

Cheers,

Andreas


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