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RE: getting the filename

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"
<msxsl:script language="JavaScript" implements-prefix="fn">
	function getFN(context){
		return context.nextNode.url;
<xsl:variable name="filename" select="fn:getFN(/)" />
<xsl:template match="/">
	<xsl:value-of select="$filename" />

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

Ciao Chris

XML/XSL Portal

>-----Original Message-----
>[]On Behalf Of Roman
>Sent: 12 July 2001 09:24
>Subject: [xsl] 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
><xsl:variable name="i" select=" ???? "/>
><form method="post">
>	<xsl:attribute name="action">
>		<xsl:value-of select="$i"/>
>	</xsl:attribute>
>best regards,
>> Roman Huditsch (RH )
>> _____________________________________________________________________
>> hico Informations- und Kommunikations-Management Gesellschaft m.b.H.
>> TechLab, Thomas A. Edison Straße 2.
>> A-7000 Eisenstadt / Austria
>> phone: +43/2682/704-61-00; fax: +43/2682/704-71-61-10
> XSL-List info and archive:

 XSL-List info and archive:

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