Hm, might it have to do with building the fragment using a key on an
external document?
I've been using sth like this:
<xsl:variable name="external" select="document('ext.xml')"/>
<xsl:variable name="fragment">
<xsl:for-each select="$fragment">
<xsl:copy-of select="key(...)"/>
</xsl:for-each>
</xsl:variable>
xalan is 2.7.1 (downloaded today)
Kind regards
P.N.
Mukul Gandhi wrote:
When I tried this sample with Xalan-J 2.7.1:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
version="1.0">
<xsl:output method="xml" indent="yes" />
<xsl:template match="/">
<xsl:variable name="fragment">
<x>
<y/>
</x>
</xsl:variable>
<xsl:variable name="node">
<node>
<xsl:copy-of select="$fragment" />
</node>
</xsl:variable>
<xsl:copy-of select="$node" />
</xsl:template>
</xsl:stylesheet>
I get the output:
<?xml version="1.0" encoding="UTF-8"?>
<node>
<x>
<y/>
</x>
</node>
So my finding does'nt match with your report ...
Could you please let us know, which version of Xalan you are using.
On Thu, Apr 17, 2008 at 7:18 PM, Peter Nabbefeld <[EMAIL PROTECTED]> wrote:
Hello!
I'm trying to convert a result tree fragment into a nodeset using
xsl:copy-of. From the w3c spec, if I understand it correct, this should be
possible:
"When a result tree fragment is copied into the result tree (see [11.3 Using
Values of Variables and Parameters with xsl:copy-of]), then all the nodes
that are children of the root node in the equivalent node-set are added in
sequence to the result tree."
I'm using the following piece of code: <xsl:variable
name="node"><node><xsl:copy-of select="$fragment"/></node></xsl:variable>
The following error occurs in this line: "#RTREEFRAG kann nicht in NodeList
konvertiert werden!"
Where's the problem?
Kind regards
Peter Nabbefeld
