[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] How to access the root node from within a subtemplate
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 -- ___________________________________________________________ Sign-up for Ads Free at Mail.com http://promo.mail.com/adsfreejump.htm
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