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[(_at_)name = 'peter']">
<xsl:call-template name="myrule">
<xsl:with-param name="nametofind" select="ref[(_at_)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[(_at_)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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