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AW: Omitting default namespace in the output - XSLT 2.0

2004-12-20 07:26:03
Thanks for the quick reply, Michael.
I hoped that I wouldn't need to use generic templates,
since that would in my case also mean many unwanted 
elements copied from my source tree. 
But having clarified that <xsl:copy-of> won't help me here, 
I think that there is no other way...

Thank you very much!

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: Michael Kay [mailto:mike(_at_)saxonica(_dot_)com] 
Gesendet: Montag, 20. Dezember 2004 15:20
An: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Betreff: RE: [xsl] Omitting default namespace in the output - XSLT 2.0

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(_dot_)Huditsch(_at_)lexisnexis(_dot_)at]
Sent: 20 December 2004 13:52
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [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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