Subject: Re: XSL-> Access to the XML Source Filename?
From: Mukul Gandhi <mukul_gandhi@xxxxxxxxx>
Date: Mon, 8 Nov 2004 08:38:13 -0800 (PST)
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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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