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Re: [xsl] XSL comparing nodesets by name only [and 1 more messages]
Subject: Re: [xsl] XSL comparing nodesets by name only [and 1 more messages] From: Piet van Oostrum <piet@xxxxxxxxxxxxxx> Date: Mon, 15 Nov 2010 08:55:16 -0400 |
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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