Translating roman numerals into integers with XSLT 2.0

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.huditsch@xxxxxxxxxxxxx
www.lexisnexis.at


> -----Urspr|ngliche Nachricht-----
> Von: Michael Kay [mailto:mike@xxxxxxxxxxxx]
> Gesendet: Montag, 20. Dezember 2004 15:20
> An: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Betreff: RE:  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.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