Martin Honnen wrote:

> >      <xsl:value-of select="name($me1) = name($me2)"/>
 > 
 > That would compare the name of the first node in $me1 with the name
 > of the first node in $me2. Any other nodes in those node sets are
 > ignored, I don't think that is what the original poster wants. I
 > think with XSLT 1.0 a single XPath expression can't solve that, a
 > template is needed.

Michael Kay wrote:

 > Unfortunately, no. In XSLT 1.0, name($nodeset) returns the name of
 > the first item in the nodeset.

Sorry, you'r right.
-- 
Piet van Oostrum
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