Hi,
 
I have a XML file that calls an XSL file for transforming to a new XML file:
 
In XML file:

<?xml version="1.0" encoding="UTF-16" standalone="no" ?>

<?xml-stylesheet href="invoices.xsl" type="text/xsl"?>

In XSL file:

<?xml version="1.0" ?>

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">

<xsl:output method="xml" indent="yes" doctype-system="test.dtd" encoding="UTF-16"/>

Transforming on a Windows2000 or NT PC give no problems.

Converting on a Windows ME PC gives following error:

javax.xml.transform.TransformerException: java.io.IOException: The system cannot find the path specified.

Code extract where error occurs:

try { // Check if xml has link to stylesheet

System.err.println("Looking for XSL file...");

Source stylesheet = tFactory.getAssociatedStylesheet(new StreamSource(xmlfile),media, title, charset);

transformer = tFactory.newTransformer(stylesheet);

System.err.println("XSL Stylesheet found...");

try {

transformer.transform(source, new javax.xml.transform.sax.SAXResult(handler));

} catch (Exception e) { //Error in transforming

System.err.println("Error in transformer.transform method...");

System.err.println("source: "+source.toString());

}

} catch (Exception e) { //No stylesheet in xml

System.err.println("No XSL stylesheet found...");

transformer = tFactory.newTransformer();

transformer.transform(source, new javax.xml.transform.sax.SAXResult(handler));

}

 
Question is which path is being looked for ?
The .xml and .xsl files are in the same directory.
 
Regards,
 
Stefan


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