AW: Omitting default namespace in the output - XSLT 2.0

I forgot to mention that the default namespace in my source document is 
prefixed with "ln".
So this explaines my match pattern "ln:table".

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(_dot_)huditsch(_at_)lexisnexis(_dot_)at
www.lexisnexis.at
 

> -----Ursprüngliche Nachricht-----
> Von: Huditsch Roman [mailto:Roman(_dot_)Huditsch(_at_)lexisnexis(_dot_)at] 
> Gesendet: Montag, 20. Dezember 2004 14:52
> An: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
> Betreff: [xsl] 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(_dot_)huditsch(_at_)lexisnexis(_dot_)at
> www.lexisnexis.at
>  
>  
>
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