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Name: <xsl:value-of select="local-name(exsl:node-set($nodes))" />
The result I expect is:
Name: nodes
But, with libxslt 1.0.22 processor, I got:
Name: fake node libxslt
To get what I want, I have to write:
Name: <xsl:value-of select="local-name(exsl:node-set($nodes)/*)" />
really tedious!!!
So, am I wrong and libxslt is right? Or viceversa ?
Thanx in advance!
Marco Guazzone
marco.guazzone@xxxxxxxxxxx
Kerbero S.r.L.
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
[xsl] exsl:node-set behaviour with variables
Subject: [xsl] exsl:node-set behaviour with variables From: sguazt <sguazt@xxxxxxxxxxx> Date: Tue, 22 Oct 2002 14:39:21 +0200 |
Hi, What happens when a value of a variable is passed to exsl:node-set() ? That is, suppose you have the following XSL fragment:
<xsl:variable name="nodes"> <nodes> <node>node 1</node> <node>node 2</node> <node>node 3</node> </nodes> </xsl:variable>
Name: <xsl:value-of select="local-name(exsl:node-set($nodes))" />
The result I expect is:
Name: nodes
But, with libxslt 1.0.22 processor, I got:
Name: fake node libxslt
To get what I want, I have to write:
Name: <xsl:value-of select="local-name(exsl:node-set($nodes)/*)" />
really tedious!!!
So, am I wrong and libxslt is right? Or viceversa ?
Thanx in advance!
Marco Guazzone
marco.guazzone@xxxxxxxxxxx
Kerbero S.r.L.
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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