[xsl] saxon 7.6.5 in SMLSpy 5 rel. 4 and xsl:result-document

Subject: [xsl] saxon 7.6.5 in SMLSpy 5 rel. 4 and xsl:result-document
From: "Kloeck, Erwin" <Erwin.Kloeck@xxxxxx>
Date: Thu, 18 Sep 2003 14:07:16 +0200
Hi,

I am using XMLSpy 5 rel. 4 and saxon 7.6.5 as XSLT Processor.
I do this with the following instruction in Tools > Options... > XSL

	java -cp D:\Programme\net\sourceforge\saxon\7.6.5\saxon7.jar net.sf.saxon.Transform -o %2 %1 %3

I am using j2sdk1.4.2.

When I process the following xsl file from within XMLSpy, I get no error message but I also
don't get the file "concat.result.xml". Everything works as expected when I call saxon from the comand line.

Anybody know what I'd doing wrong?

Thanks a lot.

Erwin

------------ concat.xsl ------------

<?xml version="1.0" encoding="UTF-8"?>
<?xmlspysamplexml D:\projects\ehecatl\web\xml\concat.xml?>
<xsl:stylesheet 
		version="2.0"
		xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
			
	<xsl:output name="xml-format" method="xml" indent="yes"/>
		
	<xsl:template match="/">
		<xsl:result-document href="concat.result.xml" format="xml-format" >
			<xsl:text disable-output-escaping="yes">
				&lt;top&gt;
			</xsl:text> 
			<xsl:for-each select="/concat/file">
				<xsl:copy-of select="document(@source)"/>
			</xsl:for-each>
			<xsl:text disable-output-escaping="yes">
				&lt;/top&gt;
			</xsl:text> 
		</xsl:result-document>
	</xsl:template>
</xsl:stylesheet>

------------ concat.xml ------------

<?xml version="1.0" encoding="UTF-8"?>
<?xml-stylesheet type="text/xsl" href="D:\projects\ehecatl\web\xml\concat.xsl"?>
<concat 
		xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance";
		xsi:noNamespaceSchemaLocation="concat.xsd">
  <file source="file1.xml"/>
  <file source="file2.xml"/>
  <file source="file3.xml"/>
</concat>





..............................

Erwin Kloeck
Produktentwicklung

Oestreicher + Wagner 
Medientechnik GmbH
Frankenthaler Strasse 20
D-81539 Muenchen

Fon   +49 (0)89-68961 216 
Fax   +49 (0)89-68961 271


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