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Re: Why do the namespace appears in transformation ?

Subject: Re: Why do the namespace appears in transformation ?
From: Michael Kay <mike@xxxxxxxxxxxx>
Date: Fri, 20 Aug 2010 09:13:56 +0100
Re:  Why do the namespace appears in transformation ?
You could get rid of the declaration by outputting a wrapper element such as xsl:stylesheet. But if you actually want the xsl:template elements to be at the top level, then the XSLT processor will always ensure that the "xsl" prefix is properly declared - the output must be well formed.

Michael Kay

On 20/08/2010 07:53, Fabien Tillier wrote:
Hi List.
I am trying to generate an XSL template from an XML file that describes
some filters.
So, basically, I get
       <TA_TITRE_A>Titre 1</TA_TITRE_A>
       <TA_TITRE_A>Titre 2</TA_TITRE_A>

I am parsing it with (not finished, of course)

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="2.0"
  <xsl:output method = "xml" encoding="UTF-8"/>
  <!--  When transforming, all xso namespace elements will become xsl -->
  <xsl:namespace-alias stylesheet-prefix="xso" result-prefix="xsl"/>
  <xsl:template match="/Results">
  	<xsl:for-each-group select="Row" group-by="MTF_NUMERO_TABLEAU">
  		<xsl:variable name="numtableau">
  			<xsl:value-of select="current-grouping-key()"/>
  		<xso:template match="node()[(MTF_NUMERO_TABLEAU =
  			<!--Do something-->
		<xsl:value-of select="current-grouping-key()"/>,

  <xsl:template match="Row">

The output (with Kernow) is

<?xml version="1.0" encoding="UTF-8"?>
<xsl:template xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
match="node()[(MTF_NUMERO_TABLEAU = '2')]"/>2,		
<xsl:template xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
match="node()[(MTF_NUMERO_TABLEAU = '3')]"/>3

Which is almost what I want....
But, how can I get rid of the xmlns declaration in each<xsl:template
section ?
(So that to get
<?xml version="1.0" encoding="UTF-8"?>
<xsl:template match="node()[(MTF_NUMERO_TABLEAU = '2')]"/>2,		
<xsl:template match="node()[(MTF_NUMERO_TABLEAU = '3')]"/>3

Thanks in advance

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