<xsl:copy> copies an element without change. That means its namespace in the
result document will be the same as its namespace in the source document (in
this case, the null namespace). To change the namespace of an element, use
<xsl:element name="{local-name()}" namespace="..."/>.
Michael Kay
http://www.saxonica.com/
-----Original Message-----
From: christoph(_dot_)klocker(_at_)rcm(_dot_)at
[mailto:christoph(_dot_)klocker(_at_)rcm(_dot_)at]
Sent: 12 June 2006 07:16
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] root siblings contain xmlns=""
Hi,
I do an transformation where I add a a schema declaration to the root
element:
in the Stylesheet I define the root element as follows:
<xsl:template match=3D"Element">
<XML xmlns=3D"http://www.something.at/Something"; xmlns:xsi=3D"http:
//www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation=3D"http:
//www.something.at/Something
http://www.something.at/something.xsd";>
<xsl:apply-templates/>
<Arranger>
<xsl:apply-templates
select=3D"$Var//ArrangerList/*"/>
</Arranger>
</XML>
<xsl:template match=3D"node()|@*">
<xsl:copy>
<xsl:apply-templates/>
</xsl:copy>
</xsl:template>
in the result document all the siblings of the root Element
contain the empty namespace ' <element xmlns=3D""> '
How can I get rid of this=3F=3F
I am using Saxon8B
thx Christoph
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