Re: XSL-> Access to the XML Source Filename?
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.. __________________________________ Do you Yahoo!? Check out the new Yahoo! Front Page. www.yahoo.com
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