Obviously xslt only knows the local tree at the current recursion level.
Is there a way to access the root node of the whole outer tree?

Assume the following (simplified) tree:

<names>
  <myelem name="paul">
    <ref name="karl" />
  </myelem>
  <myelem name="peter">
    <ref name="karl" />
  </myelem>
  <myelem name="karl">
    <ref name="tony" />
  </myelem>
</names>

The (simplified) XSLT stylesheet looks like:

....
<xsl:template match="//myelem[@name = 'peter']">
  <xsl:call-template name="myrule">
     <xsl:with-param name="nametofind" select="ref[@name]" />
  </xsl:call-template>
</xsl:template>
...

<xsl:template name="myrule">
  <!-- at the first call $nametofind contains 'karl' -->
  <xsl:param name="nametofind" />
  <xsl:call-template name="myrule">
     <!-- now I want to jump to the node <myelem name="karl"> with the subnode
'tony'
          but this doesn't work because this node is outside the context/scope
-->
     <xsl:with-param name="nametofind" select="//myelem[@name = $nametofind]"
/>
  </xsl:call-template>

</xsl:template>


How do I access from a given subtree the outer whole tree resp. other nodes?

Thank you
Matt

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