Hi,
i am doing an xsl transformation
i have my xml,xsl and the java class in same folder when i run this code from java class its working fine but when use the same in a webapplication deployed in tomcat i am getting the below error.

CODE:

StreamSource source = new StreamSource("abc.xml")
StreamSource stylesource = new StreamSource("abc.xsl"); TransformerFactory factory = TransformerFactory.newInstance();
Transformer transformer = factory.newTransformer(stylesource); StreamResult result = new StreamResult(System.out);

 StringWriter sw = new StringWriter();
transformer.transform(source, new StreamResult(sw));

String sReturn = sw.toString();


Exception:
ERROR: 'C:\Documents and Settings\hp.ln\Desktop\apache-tomcat-6.0.16\bin\abc.xsl (The system cannot find the file specified)'
FATAL ERROR: 'Could not compile stylesheet'
javax.xml.transform.TransformerConfigurationExcept ion: Could not compile stylesheet
    at com.sun.org.apache.xalan.internal.xsltc.trax.Trans formerFactoryImpl.newTemplates(Unknown Source)
    at com.sun.org.apache.xalan.internal.xsltc.trax.Trans formerFactoryImpl.newTransformer(Unknown Source)


pls anyone sort out my prob .Thanks in ADV---Raj

new StreamSource("abc.xml")

It's not a good idea to use a relative URI as an argument to "new StreamSource()". The specification doesn't say what it's relative to, and therefore the effect is completely unpredictable. It will probably be the current directory, but that's not a very predictable concept either in an environment like tomcat. Use the mechanisms you would use to access any other file in a web server environment, for example getResourceAsStream.

Michael Kay
http://www.saxonica.com/
Author, XSLT 2.0 and XPath 2.0 Programmer's Reference

Thanks Kay

String sXSLFileName = "directory.xsl";

InputStream is = getClass().getResourceAsStream(sXSLFileName);

i have used this way and i am done with it.