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
Cochabamba. URL: http://pietvanoostrum.com/
Nu Fair Trade woonartikelen op http://www.zylja.com
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