does java fully implement XSL processor?

Hello,
I am using xsl and java for XSL processing, I am currently using java sun
1.4.2 to transform a XML and XSL to XML.In the XSL I am using the document(uri)
function to retrieve documents, this function works if the uri is in this
folder, but since I need it to take a http uri, it throws me back this error



file:///C:/Projects/doctest.xsl; Line #6; Column #-1; C
an not load requested doc: Expected "</META>" to terminate element starting
on l
ine 3.

I have testing this against Microsoft's compiler and it worked grand so
I am left wondering whats the craic with java, this function worked if I
replaced the http uri with say a file in my folder like 'web.xml'

Here is the XSL code

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:fo="http://www.w3.org/1999/XSL/Format">
<xsl:template match="/xslviewjsp">
<doctest>
<info>
<xsl:variable name="foo" select="document('http://localhost:8080/exist/xmldb/db/noel_collection/xslview_web/web2.xml')"/>
<xsl:variable name="doc1" select="document('web2.xml')"/>
<xsl:variable name="doc2" select="document('web.xml')"/>
<xsl:if test="$doc1/xslviewjsp/title/text() = $doc2/xslviewjsp/title/text()">
<xsl:for-each select="document('web2.xml')/xslviewjsp/*">
<xsl:element name="{name()}">
<xsl:value-of select="."/>
</xsl:element>
</xsl:for-each>
</xsl:if>
</info>
</doctest>
</xsl:template>
</xsl:stylesheet>

The class I am using to transform the xsl is java's standard class Transformer

Hope somebody can help me
Thanks
Noel 


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