Re: Why do the namespace appears in transformation ?

Michael Kay
Saxonica

Hi List.
I am trying to generate an XSL template from an XML file that describes
some filters.
So, basically, I get
<Results>
    <Row>
       <MTF_NUMERO_TABLEAU>2</MTF_NUMERO_TABLEAU>
       <TA_TITRE_A>Titre 1</TA_TITRE_A>
    </Row>
    <Row>
       <MTF_NUMERO_TABLEAU>3</MTF_NUMERO_TABLEAU>
       <TA_TITRE_A>Titre 2</TA_TITRE_A>
    </Row>
  </Results>

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

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="2.0"
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
exclude-result-prefixes="xsl"
  xmlns:xso="dummy"
   >
  <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()"/>
  		</xsl:variable>
  		<xso:template match="node()[(MTF_NUMERO_TABLEAU =
'{$numtableau}')]">
  			<!--Do something-->
  		</xso:template>
		<xsl:value-of select="current-grouping-key()"/>,

	</xsl:for-each-group>
  </xsl:template>
  <xsl:template match="Row">
  	
  </xsl:template>
</xsl:stylesheet>

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
Regards,
Fabien