RE: [xsl] Transforming XML to XML

Subject: RE: [xsl] Transforming XML to XML
From: "Pilarski, James" <James_Pilarski@xxxxxxxxxxxxxxxx>
Date: Wed, 11 Aug 2004 17:11:28 -0400
Per your suggestion, I have tried both of the following stylesheets:

<?xml version="1.0" encoding="ISO-8859-1"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
<xsl:output method="xml" version="1.0" encoding="iso-8859-1" indent="yes"/>

<xsl:template match="/">
  <xsl:apply-templates select="Data"/>
</xsl:template>

<xsl:template match="Data">
  <xsl:copy-of select="."/>
</xsl:template>

</xsl:stylesheet>

AND

<?xml version="1.0" encoding="ISO-8859-1"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
<xsl:output method="xml" version="1.0" encoding="iso-8859-1" indent="yes"/>

<xsl:template match="/">
  <xsl:copy-of select="Data"/>
</xsl:template>

</xsl:stylesheet>

Both of these stylesheets are still returning just a string from the following
XML:

<?xml-stylesheet type="text/xsl" href="renewal.xsl"?>
<Data>
  <car>Chevy</car>
  <car>Dodge</car>
  <car>Ford</car>
</Data>

Are the modifications that I made to the example above incorrect?
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