Hi again,
Sorry, for bothering you a second time today,
but I was wondering if there is a more or less simple way
of translating roman numerals into intergers with XSLT 2.0.
Or do I have to use the standard method for doing this
describes in the "XSLT Cookbook" by Sal Mangano?
Would be nice if format-number() or anything similar would allow
me to o this...
Thank you very much for your patience.
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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