Subject: [xsl] javax.xml.transform.TransformerException: xsl:stylesheet is not a llowed in this position in the stylesheet From: Charles Macleod <cmacleod@xxxxxxxx> Date: Thu, 6 May 2004 12:17:06 -0700 |
All, I am using the Xalan-J TemplatesHandler via javax.xml.transform.sax.TemplatesHandler. I am (now) parsing an 'empty'xsl (as a test) with an XMLReader that has the TemplatesHandler set as its ContentHandler (see code below). I keep getting a "javax.xml.transform.TransformerException: xsl:stylesheet is not allowed in this position in the stylesheet!" error. I stripped everything out of the xsl leaving only the <xsl:stylesheet> element and I still get the error. Any ideas? I'm figuring this has to be a configuration issue with the reader but I really don't have a clue.... Stylesheet: <?xml version="1.0" encoding="UTF-8"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:fo="http://www.w3.org/1999/XSL/Format"> </xsl:stylesheet> Code: TransformerFactory tfac = TransformerFactory.newInstance() ; SAXTransformerFactory saxTfac = ((SAXTransformerFactory) tfac) ; TemplatesHandler templatesHandler = saxTfac.newTemplatesHandler() ; // Create an XMLReader and set its ContentHandler. SAXParserFactory fac = SAXParserFactory.newInstance() ; SAXParser parser = fac.newSAXParser() ; XMLReader reader = parser.getXMLReader() ; reader.setContentHandler( templatesHandler) ; // Parse the stylesheet. reader.parse( args[0]) ; <--- throws exception
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