Hi John !
<xsl:value-of select="$somenode/@attr" /> is not correct i think, the select
attribute shall be wheter an variable name precede with or an Xpath
expression.
<xsl:copy-of select="/root/node1" /> will actually return the xml fragment
of node1, but when doing <xsl:value-of select=""/> you will get the value of
this fragment.
I'm not sure I really got what you mean by "return a reference to a node" ?
An idea (maybe) is to use the generate-id() Xpath function which gives you
an single id  for each node ?
hope this help.
Matthieu

-----Message d'origine-----
De : John [mailto:john-xsl-list@xxxxxxxx]
Envoyi : jeudi 16 juin 2005 16:25
@ : xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Objet :  Can a named template return a node list?


Sorry in advance for my terminology.  Is it possible for a named
template to return a node list?  Here is some simplified XML with which
I have been experimenting:

<root>
   <node1 attr="node1attrval">
     <node1a>node1atext</node1a>
   </node1>
   <node2 attr="node2attrval">
     <node2a>node2atext</node2a>
   </node2>
</root>

And the XSL I have been trying:

<xsl:template match="*">
   <xsl:variable name="somenode">
     <xsl:call-template name="sometemplate" />
   </xsl:variable>
   <xsl:value-of select="$somenode/@attr" />
</xsl:template>

<xsl:template name="sometemplate">
   <xsl:copy-of select="/root/node1" />
</xsl:template>

The reference to $somenode/@attr gives an error.  I think the problem is
with copy-of - I want it to return a reference to a node, but it seems
to return the value of the node.  Is there some other function or
approach I should try?  I have tried value-of and copy, and I am hoping
to stay away from XSL extensions if possible.  I cannot use XSL 2.

Thanks,

    -John