[xsl] XSL to XSL stylesheet: namespace question

Hello xsl-list,

  could anyone give me an insight on how to make correct transform
  from xsl to xsl.
  Here is an example (the bad thing is that it makes ns0:stylesheet
  xmlns:ns0="http://www.w3.org/1999/XSL/Transform"; and thus xsl goes
  as undeclared ns):

  source xsl:
<?xml version="1.0" encoding="UTF-8"?>
<?xml-stylesheet type="text/xsl" href="design.xsl"?>
<adt:root xmlns:adt="http://somewhere.domain/ADT"; xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
<html>
Some here
</html>
</adt:root>
  
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"; xmlns:adt="http://somewhere.domain/ADT";>
<xsl:output method="xml" indent="yes" version="1.0" encoding="UTF-8"/>
<xsl:template match="/">
<xsl:apply-templates/>
</xsl:template>
<xsl:template match="adt:root">
<xsl:element name="xsl:stylesheet" namespace="http://www.w3.org/1999/XSL/Transform";>
<xsl:attribute name="version">1.0</xsl:attribute>
<xsl:element name="xsl:output">
<xsl:attribute name="method">xml</xsl:attribute>
<xsl:attribute name="version">1.0</xsl:attribute>
<xsl:attribute name="encoding">UTF-8</xsl:attribute>
<xsl:attribute name="indent">yes</xsl:attribute>
</xsl:element>
<xsl:apply-templates/>
</xsl:element>
</xsl:template>

<xsl:template match="*|@*">
<xsl:copy>
<xsl:apply-templates select="@*|node()"/>
</xsl:copy>
</xsl:template>

</xsl:stylesheet>

-- 
Best regards,
 viewga                          mailto:viewga@xxxxxxxxxxxx




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