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RE: [xsl] Extract input filename

2007-05-17 01:44:29
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
>
>
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