RE: getting the filename

Roman,
If you are doing this in the browser (with linked stylesheet) which it seems
that you are then you can't pass this info in as a parameter. You need to do
something like
<?xml version="1.0" encoding="utf-8" ?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:msxsl="urn:schemas-microsoft-com:xslt"
xmlns:fn="cjb"
>
<msxsl:script language="JavaScript" implements-prefix="fn">
	function getFN(context){
		return context.nextNode.url;
	}
</msxsl:script>
<xsl:variable name="filename" select="fn:getFN(/)" />
<xsl:template match="/">
	<xsl:value-of select="$filename" />
</xsl:template>
</xsl:stylesheet>

<?xml version="1.0" encoding="utf-8" ?>
<?xml-stylesheet type="text/xsl" href="test.xsl"?>
<test />

Ciao Chris

XML/XSL Portal
http://www.bayes.co.uk/xml


>-----Original Message-----
>From: owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx
>[mailto:owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx]On Behalf Of Roman
>Huditsch
>Sent: 12 July 2001 09:24
>To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
>Subject:  getting the filename
>
>
>Good morning everybody!
>
>I have a slight problem with my xsl-stylesheet. I have to store the
>filename of my XML-file, which calls the xsl-stylesheet, in a variable
>within my XSL-file. Is there a way to get that filename into my
>variable?
>
>
><xsl:variable name="i" select=" ???? "/>
>
><form method="post">
>	<xsl:attribute name="action">
>		<xsl:value-of select="$i"/>
>	</xsl:attribute>
></form>
>
>
>best regards,
>Roman
>
>> Roman Huditsch (RH )
>> _____________________________________________________________________
>> hico Informations- und Kommunikations-Management Gesellschaft m.b.H.
>> TechLab, Thomas A. Edison Straße 2.
>> A-7000 Eisenstadt / Austria
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>> e-mail:support@xxxxxxxx;  r.huditsch@xxxxxxxx
>>
>>
>
> XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
>
>


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