Subject: Re: [xsl] RE: syntax sugar for call-template From: David Carlisle <davidc@xxxxxxxxx> Date: Fri, 16 Feb 2001 12:45:31 GMT |
I've been - perhaps lazily - assuming that the implicit RTF -> nodeset conversion would convert any node structure into the same structure you started with. For instance if I return a node-set of node-sets I'm expecting this to be what the calling expression receives. Do you know of any exceptions, or are you just being cautious? You can't get nodes from source documents into an node set generated from an rtf (or in XSL 1.1 terminology) generated by an xsl:variable with non empty content. <xsl:variable name="beers" select="beer"/> gives you a node set of all beer children of the current node. <xsl:variable name="beers"> <xsl:copy-of select="beer"/> </xsl:variable> gives you an rtf in 1.0 but if you saxon:node-set() that, or use 1.1 you get a node set with _one_ node (a root node) containing a bunch of beer nodes but they are _copies_ of the original nodes, not the same nodes (so they have different generate-id() values, different parents etc). parent::beer in the first case will give you the original current node in teh second case it will give the root node of the document. There is no way for a named template to emulate copy-of which is one reeasons why viewing XSLT functions as syntaxtic sugar for named templates may not be quite the right thing and you need something like saxon:return. David _____________________________________________________________________ This message has been checked for all known viruses by Star Internet delivered through the MessageLabs Virus Control Centre. For further information visit http://www.star.net.uk/stats.asp XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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