RE: [xsl] Problems doing XSL Transformation using ASP

Subject: RE: [xsl] Problems doing XSL Transformation using ASP
From: Mukul Gandhi <mukulgw3@xxxxxxxxx>
Date: Sat, 11 Oct 2003 05:02:07 -0700 (PDT)
Sorry I did'nt knew Msxml2 is MSXML3. I applied the
XSL to XML *without changing anything* with MSXML3
(not through ASP but with xmlspy) on Win2K. The
transformation succedded. 

I am curious, if you know the answer -- What does the
number "2" signify in Msxml2.DomDocument ?

Regards,
Mukul


--- Americo_Albuquerque <melinor@xxxxxxxx> wrote:

> No Rangarajan is using MSXML3. Msxml2.DomDocument is
> MSXML3. 
> 
> The error reported says: msxml3.dll error '80004005'
> Also an evidance that he is using MSXML3
> 
> (...)
> 
> > > 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???
> 
> Check if your dummy.xsl is in the same directory as
> the asp file
> Open the dummy.xsl in IE to see if it is well formed
> 
> > > 
> > > 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");
> Change this to:
> xslDoc.load(Server.MapPath("dummy.xsl"));
> Sometimes asp as a little trouble in finding the
> files
> 
> Add here:
>        if(xslDoc.parseError.errorCode!=0) {
>          response.write("Stylesheet error:<br>");
>          response.write("Error: " +
> xslDoc.parseError.errorCode +
> "<br>");
>          response.write("Reason: " +
> xslDoc.parseError.reason + "<br>");
>          response.write("Line: " +
> xslDoc.parseError.line + "<br>");
>          response.write("Char position: " +
> xslDoc.parseError.linepos +
> "<br>");
>          response.write("Source: " +
> xslDoc.parseError.srcText +
> "<br>");
>        }
> 
> This will report your error with more accuracy
> 
> > >    xslt.stylesheet = xslDoc;
> > >    var xmlDoc = new
> > > ActiveXObject("Msxml2.DOMDocument");
> > >    xmlDoc.async = false;
> > >    xmlDoc.resolveExternals = false;
> > >    xmlDoc.load("dummy.xml");
> 
> You might want to do the same here. Just copy the
> above code and change
> xslDoc by xmlDoc
> 
> > >    xslProc = xslt.createProcessor();
> > >    xslProc.input = xmlDoc;
> > >    xslProc.transform();
> > >    Response.Write(xslProc.output)
> > > %>
> (...)
> 
> Regards,
> Americo Albuquerque
> 
> 
>  XSL-List info and archive: 
> http://www.mulberrytech.com/xsl/xsl-list
> 


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