Re: [xsl] XSL-> Access to the XML Source Filename?

[Mukul Gandhi <mukul_gandhi@xxxxxxxxx>]

With XSLT 1.0 , I believe you cannot. You could pass
the XML file name as parameter to the processor with
1.0 ..

With XSLT 2.0, I think you can do it with document-uri
function .. I just tried with Saxon 8 and it works ..

The XSL is -
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
version="2.0">

<xsl:output method="text"/>
   
<xsl:template match="/">
    <xsl:value-of select="document-uri(.)" />
</xsl:template>

</xsl:stylesheet>

When it is applied to a XML file xmlfile.xml ,
it returns result like
file:/D:/xml/xsl/xmlfile.xml

Regards,
Mukul

--- news@xxxxxxxxxxx wrote:

> is there any way to get access to the name of the
> XML file I'm currently
> tranferring?

I hope you mean transforming..



		
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