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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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