Fabien,
your output is non-XML since it does not contain a single root element.
Just adding "<result>" before "<xsl:for-each-group>" and "</result>"
after "</xsl:for-each-group>" gives below output.
The "xsl" namespace prefix needs to be defined somewhere if you use it.
Is this what you want?
$ saxon xfrom2.xsl data.xml | tidy -q -xml
<?xml version="1.0" encoding="utf-8"?>
<result xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="node()[(MTF_NUMERO_TABLEAU = '2')]" />
2,
<xsl:template match="node()[(MTF_NUMERO_TABLEAU = '3')]" />
3,</result>
$
Mit besten Gruessen / Best wishes,
Hermann Stamm-Wilbrandt
Developer, XML Compiler, L3
WebSphere DataPower SOA Appliances
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From: "Fabien Tillier" <f(_dot_)tillier(_at_)cerep(_dot_)fr>
To: <xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com>
Date: 08/20/2010 08:52 AM
Subject: [xsl] Why do the namespace appears in transformation ?
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
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