[xsl] How can I see the resulting XML structure after a transformation f

I'm trying to perform a 2 step transformation in my JSP file. I have a XML 
structure for a document where all links are under <document_links>, the XML 
also contains <paragraph> nodes.

I want to transform this XML document so that all nodes except <paragraph> 
nodes are in the transformed XML.

I want to test if transforming from one XML structure to another was done 
correctly before using the output of the transformation. 

But there doesn't seem to be a way in JSLT to see the xml node structure. I 
tried printing it with <x:out but it only outputs the text values and not the 
xml nodes themselves.

I also tried to see the output of the transformation by performing clien-side 
transformation with IE6.0, but this transformation also shows only the text 
nodes.

How can I see the *transformed XML structure* ?

For example: 

I want to transform the following XML structure 
document.xml

<?xml version="1.0" encoding="UTF-8"?>
<document>
<!--
   Commented out , used for IE6.0 transformation.
   <?xml-stylesheet type="text/xsl" href="get_links.xsl"?>
-->
    <document_links>
        <link>
            <id>1234</id>
            <url><![CDATA[http://www.test.com]]></url>
        </link>
    </document_links>
    <paragraph>
      <![CDATA[
      test text
      ]]>
    </paragraph>
</document>

to this structure: (just trying to extract the XML with the link nodes, 
excluding paragraph nodes)
<?xml version="1.0" encoding="UTF-8"?>
<document>
<!--
   Commented out , used for IE6.0 transformation.
   <?xml-stylesheet type="text/xsl" href="get_links.xsl"?>
-->
    <document_links>
        <link>
            <id>1234</id>
            <url><![CDATA[http://www.test.com]]></url>
        </link>
    </document_links>
</document>

This is the XSL:
get_links.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" omit-xml-declaration="no" indent="no"/>
    <xsl:template match="/">
        <xsl:apply-templates select="document"/>
    </xsl
nt">
       <document>
           <xsl:apply-templates select="document_links"/>
       </document>
    </xsl:template>
    <xsl:template match="document_links">
        <document-links>
            <xsl:copy-of select="."/>
        </document-links>
    </xsl:template>
</xsl:stylesheet>

Snippet from JSP:

<c:import url="/document.xml" var="xml"/>
<c:import url="/get_links.xsl" var="xsl"/>
<x:transform xml="${xml}" xslt="${xsl}" var="links_xml"/>
<x:out select="$links_xml"/>

The above <x:out only shows 1234 http://www.test.com  and not the XML structure.

I'm not sure if this is the right way to test the output of XML transformation

Any help is appreciated.

Regards
Rashmi



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