RE: [xsl] Extract input filename

["Michael Kay" <mike@xxxxxxxxxxxx>]

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