Well acutally that would be quite a mess, but I think I've come to
a practicable solution.
Thanks for your help.
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: António Mota [mailto:xptm(_at_)sapo(_dot_)pt]
Gesendet: Montag, 20. Dezember 2004 18:30
An: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Betreff: Re: AW: [xsl] Omitting default namespace in the
output - XSLT 2.0
Write a empty template for the not wanted elements:
<xsl:template match="*">
<xsl:element name="{local-name()}">
<xsl:copy-of select="@*"/>
<xsl:apply-templates/>
</xsl:element>
</xsl:template>
<xsl:template match="not-wanted">
<xsl:template>
Quoting Huditsch Roman <Roman(_dot_)Huditsch(_at_)lexisnexis(_dot_)at>:
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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