RE: [xsl] Returning the file name of the input file

[Cams Ismael <Ismael dot Cams at siemens dot atea dot be>]

Thanks all for trying to help, but I think I havn't explained myself well
enough. What I try to do is the following:
I have an xml file which refers to other xml files that can refer to other
xml files, etc.

E.g.

	<package type="expr" href="../../package1.xml"/>
	<package href="../../package2.xml"/>
	<package type="expr" href="../../package3.xml"/>

Out of this xml file I generate a list of all references of type 'expr'. So
I search through all xml files and generate a list of all references of type
'expr'. The result file is used by a Java program to execute another
process. The problem is because relative paths are used the Java program
can't find the files (the Java program is started from another place).
Therefore I would like to translate the relative paths into absolute paths.
But I don't find any function in xsl to do this, nor do I see a solution to
solve this with xsl. Hopefully somebody can help me.

Kind regards,
Ismaël

-----Original Message-----
From: Jarno.Elovirta@nokia.com [mailto:Jarno.Elovirta@nokia.com]
Sent: Wednesday, August 28, 2002 9:57 AM
To: xsl-list@lists.mulberrytech.com
Subject: RE: [xsl] Returning the file name of the input file


Hi,

> Subject: [xsl] Returning the file name of the input file
> 
> 
> Hello,
> 
> does anybody if the following is possible:
> 
> I want to use the name of the inputfile in the outputfile. So 
> if I want to
> transform the file c:\temp\test.xml, I want to get in my output
> c:\temp\test.xml.

Pass the input filename as a parameter.

Jarno

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