Hello,

Via javascript I am loading xsl's from tomcat to client. My javascript code was 
the following...

           // sort the xml using xsl stylesheet
           xsl.load("reportwriter/xsl/treesort.xsl");
           xml.loadXML(xml.transformNode(xsl));

           // Load XSL
           xsl.load("reportwriter/xsl/tree.xsl");

This resulted in the following error...

Error: The stylesheet does not contain a document element.  The stylesheet may 
be empty, or it may not be a well-formed XML document.

I modified the code to the following

           // sort the xml using xsl stylesheet
alert(getUrl() + "/reportwriter/xsl/treesort.xsl");
           xsl.load(getUrl() + "/reportwriter/xsl/treesort.xsl");
           xml.loadXML(xml.transformNode(xsl));

           // Load XSL
alert(getUrl() + "/reportwriter/xsl/tree.xsl");
           xsl.load(getUrl() + "/reportwriter/xsl/tree.xsl");

the alert show the following

https://localhost:7443/Reporting/reportwriter/xsl/treesort.xls (I never get to 
the second alert)

If I cut/paste the url into browser url line it loads and displays without 
issue. However,  when I run my webapp I still get the following error

Error: The stylesheet does not contain a document element.  The stylesheet may 
be empty, or it may not be a well-formed XML document.

So it appears the stylesheet is loading a blank file. Any ideas? Is there 
something different I need to do with tomcat to load these?

My webapp directory structure is the following

Within tomcat webapps folder I have

Reporting
                Logs
                META-INF
                reportwriter
                                images
                                pages
                                reports
                                scripts
                                xml
                                xsl - this is where the two xsl files are stored
                WEB-INF

This is the contents of treesort.xsl

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>

<xsl:template match="menu">
                <xsl:copy>
                                <xsl:for-each select="menuItem">
                                                <xsl:sort select="@title"/>
                                                <xsl:copy>
                                                                <xsl:copy-of 
select="@*"/>
                                                                <xsl:for-each 
select="menuItem">
                                                                                
<xsl:sort select="@title"/>
                                                                                
<xsl:copy-of select="."/>
                                                                </xsl:for-each>
                                                </xsl:copy>
                                </xsl:for-each>
                </xsl:copy>
</xsl:template>

</xsl:stylesheet>

Sincerely,

Robert Jenkin
Surecomp Services, Inc.
2 Hudson Place, 4th Floor
Hoboken, NJ 07030
Skype: robert.jenkin
Office: 201 217 1437 | Direct: 201 716 1219 | Mobile: 908 251 0537
http://www.Surecomp.com


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