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Re: [xsl] Namespace on output node but not not in source

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Ok, it is in fact working, the namespace doesn't appear with
exclude-result-prefixes="xs". My mistake was that i still have my
original problem, so that means that it's not a XSL problem, so i'm
back to the drawing board...

Thanks for your help.

On 10/18/05, Michael Kay <mike@xxxxxxxxxxxx> wrote:
> The rules for a literal result element say that all its in-scope namespaces
> are copied to the result tree.
>
> However, you should be able to prevent this using exclude-result-prefixes.
> If this isn't working, please supply a complete (but small!) stylesheet to
> demonstrate the problem so we can see what you are doing wrong.
>
> Michael Kay
> http://www.saxonica.com/
>
> > -----Original Message-----
> > From: Antsnio Mota [mailto:amsmota@xxxxxxxxx]
> > Sent: 18 October 2005 00:59
> > To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> > Subject: [xsl] Namespace on output node but not not in source
> >
> > Hi:
> >
> > I have a xsl with a namespace declared
> >
> > <xsl:transform xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
> >       version="1.0"
> >       xmlns:xs="http://www.w3.org/2001/XMLSchema">
> >
> > so i can make some lookups on a xsd file
> >
> >  <xsl:variable name="tipo"
> > select="$schema/xs:schema/xs:element[@name=$nome]/@type"/>
> >
> > and that's all i do, i don't copy any node from $schema to
> > the output tree.
> >
> > However, this
> >
> >         <xsl:template match="/">
> >                 <table>
> >                         <xsl:apply-templates
> > select="(//Menu)[position()=$pos]"/>
> >                 </table>
> >         </xsl:template>
> >
> > produces a
> >
> > <table xmlns:xs="http://www.w3.org/2001/XMLSchema">
> >      <...>
> > </table>
> >
> > and i don't understand why, and i don't want it there.
> >
> > I add a exclude-result-prefixes="xs" but it seems it makes no
> > diference.
> >
> > Why is this?
> >
> >
> > Thanks.

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