Folks,
I have a very simple XML, XSL file and a very simple ASP file to perform the
XSL Transformation on the XML file using the XSL file and providing it to
any client browswer. I get the following error.
NOTE: If I open the xml file (that is just from the client side) it works
just fine!
What am I missing???
ERROR
---------
msxml3.dll error '80004005'
The stylesheet does not contain a document element. The stylesheet may be
empty, or it may not be a well-formed XML document.
/test/vqtreport.asp, line 9
ASP FILE - dummy.asp
------------
<%@ language=javascript %>
<%
var xslt = new ActiveXObject("Msxml2.XSLTemplate");
var xslDoc = new ActiveXObject("Msxml2.FreeThreadedDOMDocument");
var xslProc;
xslDoc.async = false;
xslDoc.resolveExternals = false;
xslDoc.load("dummy.xsl");
xslt.stylesheet = xslDoc;
var xmlDoc = new ActiveXObject("Msxml2.DOMDocument");
xmlDoc.async = false;
xmlDoc.resolveExternals = false;
xmlDoc.load("dummy.xml");
xslProc = xslt.createProcessor();
xslProc.input = xmlDoc;
xslProc.transform();
Response.Write(xslProc.output)
%>
XML FILE - dummy.xml
------------
<?xml version="1.0"?>
<?xml-stylesheet type="text/xsl" href="dummy.xsl"?>
<DUMMY>
</DUMMY>
XSL FILE - dummy.xsl
------------
<?xml version='1.0'?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
<xsl:output method="html" indent="yes"/>
<xsl:template match="/">
<html>
<h1>Yahoooooooo</h1>
</html>
</xsl:template>
</xsl:stylesheet>
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