RE: Omitting default namespace in the output - XSLT 2.

If you copy elements using xsl:copy-of, the output elements will have the
same names as the input elements, and the system will automatically add
declarations of the namespaces used in these names. You want your output
elements to have a different name from the input elements (same local name,
different URI). So you can't use copy-of.

To copy a tree while renaming elements, use a modified form of the identity
template rule:

<xsl:template match="*">
<xsl:element name="{local-name()}">
  <xsl:copy-of select="@*"/>
  <xsl:apply-templates/>
</xsl:element>
</xsl:template>

Michael Kay
http://www.saxonica.com/
 

> -----Original Message-----
> From: Huditsch Roman [mailto:Roman.Huditsch@xxxxxxxxxxxxx] 
> Sent: 20 December 2004 13:52
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Subject:  Omitting default namespace in the output - XSLT 2.0
> 
> Hi,
> 
> Given an example input like:
> 
> <?xml version="1.0" encoding="UTF-8"?>
> <norm xmlns="myDefaultNamespace">
> 	<table>
> 		<row>
> 			<cell>My Table</cell>
> 		</row>
> 	</table>
> </norm>
> 
> I searched for an easy way to get output data, 
> which is not associated to my default namespace any more,
> with the help of <xsl:copy-of> in XSLT 2.0
> I hoped that the attribute "copy-namespaces" set to "no" would help me
> here,
> but unfortunately I had no luck with Saxon 8.1.1
> 
> My output still looks like 
> 
> <table>
> 	<row xmlns="myDefaultNamespace">
> 		<cell>My Table</cell>
> 	</row>
> </table>
> 
> 
> XSLT:
> =====
> 
> 	<xsl:template match="ln:table">
> 		<table>
> 			<xsl:copy-of select="node() | @*"
> copy-namespaces="no"/>
> 		</table>
> 	</xsl:template>
> 
> 
> Thanks in advance for your input!
> 
> wbr,
> Roman
> _______________________________________
> 
> Roman Huditsch
> IT and Electronic Publishing
> LexisNexis ARD Orac 
> Marxergasse 25
> 1030 Vienna
> Austria 
> ph: +43-1-534 52-1514
> f: +43-1-534 52-140
> e-mail roman.huditsch@xxxxxxxxxxxxx
> www.lexisnexis.at