XSLT: I don't get the xmlns in target

[Rosa Mª Gómez Flores]

Hallo!

I don't understand exactly what is the 'xmlns' attribute, nor why it have
problems with the xmlns attribute of the sylesheet, but I need it in
my output XML file. Could anybody tell me how to transform this input:

<ROOT-ELEMENT>
<SOHN>...</SOHN>
</ROOT-ELEMENT>

in this output:

<ROOT-ELEMENT xmlns="http://www.whatever">
<SOHN>...</SOHN>
</ROOT-ELEMENT>

Thanks a lot of,

Rgf

 

[Martin Boehm]

Could anybody tell me how to transform this input:

<ROOT-ELEMENT>
<SOHN>...</SOHN>
</ROOT-ELEMENT>

in this output:

<ROOT-ELEMENT xmlns="http://www.whatever">
<SOHN>...</SOHN>
</ROOT-ELEMENT>

<?xml version='1.0'?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/ROOT-ELEMENT">
<xsl:element name="ROOT-ELEMENT" namespace="http://www.whatever">
<xsl:copy-of select="./*"/>
</xsl:element>
</xsl:template>
</xsl:stylesheet>

Martin

 

[Dimitre Novatchev]

This is not the correct solution -- the copied nodes will not belong to the
namespace of ROOT-ELEMENT.

=====
Cheers,

Dimitre Novatchev.
http://fxsl.sourceforge.net/ -- the home of FXSL

 

[Richard Tobin]

I don't understand exactly what is the 'xmlns' attribute, nor why it have
problems with the xmlns attribute of the sylesheet,

You might want to find out more before blindly pressing ahead, but
this copies a document, placing all elements in the http://www.whatever
namespace:

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">

<xsl:template match="*">
<xsl:element name="{local-name()}" namespace="http://www.whatever">
<xsl:apply-templates select="node()|@*"/>
</xsl:element>
</xsl:template>

<xsl:template match="@*|comment()|processing-instruction()|text()">
<xsl:copy/>
</xsl:template>

</xsl:stylesheet>

-- Richard

 

[Bob Foster]

This is not the correct solution -- the copied nodes will not belong to the
namespace of ROOT-ELEMENT.

Here's a better one:

<xs:stylesheet xmlns:xs="http://www.w3.org/1999/XSL/Transform"
version="1.0">
<xsutput method="xml" omit-xml-declaration="yes"/>
<xs:template match="*" priority="1.0">
<xs:element name="{name(.)}" namespace="http://example.com">
<xs:apply-templates select="@*|node()"/>
</xs:element>
</xs:template>
<xs:template match="@*|node()">
<xs:copy>
<xs:apply-templates select="@*|node()"/>
</xs:copy>
</xs:template>
</xs:stylesheet>

Bob Foster

 

[Dimitre Novatchev]

Here's a better one:

Yes, I already provided exactly the same in microsoft.public.xsl -- as I
pointed to Rosa, no one would reply to her identical messages in 3 forums.


In microsoft.public.xml Marrow posted a slightly better one -- the default
namespace is declared on the xsl:stylesheet element and does not need to be
specified on the xsl:element instructions.


=====
Cheers,

Dimitre Novatchev.
http://fxsl.sourceforge.net/ -- the home of FXSL

 

[Bob Foster]

In microsoft.public.xml Marrow posted a slightly better one -- the default
namespace is declared on the xsl:stylesheet element and does not need to be
specified on the xsl:element instructions.

I don't know how to do that. What does it look like?

Bob

 

[Dimitre Novatchev]

http://groups.google.com/groups?dq=...8&oe=UTF-8&q=microsoft.public.xml&sa=N&tab=wg


=====
Cheers,

Dimitre Novatchev.
http://fxsl.sourceforge.net/ -- the home of FXSL