RE: [xsl] Extract input filename

["J. S. Rawat" <jrawat@xxxxxxxxxxxxxx>]

Thanks for the prompt reply. Would you please let me know about the 
namesspace value of prefix {xs} as an error is throughing while converting.

Error
undeclared namespace prefix {xs}

At 08:46 AM 5/17/2007 +0100, you wrote:

> 1. Change this:
>
> <xsl:variable name="filename" select="document('input_meta.xml')" />
>
> to this:
>
> <xsl:param name="meta" as="xs:string" required="yes"/>
> <xsl:variable name="filename" select="document($meta)" />
>
> (you might also like to change the name of the variable filename, as it's a
> misleading name)
>
> 2. Change your command line to
>
> java -jar saxon8.jar input.xml x.xsl meta=input_meta.xml >c.xml
>
> Michael Kay
> http://www.saxonica.com/
>
>
> > I have no idea about to extract input filename by passing
> > argument in command line. In the below examples, I don't want
> > to hardcode "input_meta.xml".
> >
> > Input files
> > 1. input.xml
> > 2. input_meta.xml
> >
> > command line
> > java -jar saxon8.jar input.xml x.xsl >c.xml
> >
> > Stylesheet
> > <xsl:variable name="filename" select="document('input_meta.xml')" />
> >  <xsl:template match="/">
> >       <doi>
> >        <xsl:copy-of select="$filename/doi/text()" />
> >       </doi>
> > </xsl:template>
> >
> > Thanks in advance.
> > JSR