RE: [xsl] Translating roman numerals into integers with XSLT 2.0

Subject: RE: [xsl] Translating roman numerals into integers with XSLT 2.0
From: Andrew Curry <andrew.curry@xxxxxxxxxxxx>
Date: Mon, 20 Dec 2004 16:31:49 -0000
i would do it by defining the numbers in a seperate xml document for each
roman numeral and in the xsl process the numeral character by character
getting the number from the document and adding it. Not sure this is what
the cookbook suggests.

-----Original Message-----
From: Huditsch Roman [mailto:Roman.Huditsch@xxxxxxxxxxxxx]
Sent: 20 December 2004 16:20
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: [xsl] 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: [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.Huditsch@xxxxxxxxxxxxx]
> > Sent: 20 December 2004 13:52
> > To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> > 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.huditsch@xxxxxxxxxxxxx
> > www.lexisnexis.at

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