Using Style Sheet w/ XSQL

I have been getting into XSQL and hit a wall when trying to
display my XSLQ w/ a 2nd style sheet.
This is what I have been doing:
1)XSQL query.
File Name (d2.xsql)
2)I then re-format that query w/ a stylesheet and return XML.
FILE NAME (d3.xsql)
<?xml version="1.0"?>
<?xml-stylesheet type="text/xsl" href="devcorner.xsl"?>
<xsql:include-xsql href="d2.xsql" xmlns:xsql="urn:oracle-xsql"/>
3) Now I want to display my XML document dynamically w/ a
stylesheet and output XML
<?xml version="1.0"?>
<?xml-stylesheet type="text/xsl" href="devcorner.xsl"?>
<xsql:include-xsql href="d3.xsql" xmlns:xsql="urn:oracle-xsql"/>
However step 3 always returns an error of:
-->XML document must have a top level element.
Any ideas on how I can get around this?
Thanks!
-Stephen

Shaun, thanks for the response.
The 2nd XML does have a root element (devcorner) and if I save
this XML to my hard drive I can do a transformation on it, using
my stylesheet.
This is what the 2nd XML looks like
<?xml version="1.0" encoding="UTF-8" ?>
- <devcorner>
+ <category> (collapsed)
</devcorner>
The problem has to do w/ when I try to use the following XSQL to
transform the above XML
<?xml version="1.0"?>
<?xml-stylesheet type="text/xsl" href="devcorner.xsl"?>
<xsql:include-xsql href="d2.xsql" xmlns:xsql="urn:oracle-xsql"/>
thanks.
-Stephen

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