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

[Cams Ismael <Ismael.Cams@xxxxxxxxxxxxxxx>]

That would be a possibility if I only have to transform one file. But what I
use now is an xml file that refers to other xml files. With the document()
function I traverse all these xml files. By passing a parameter to the
source file I have only the name of one file. What I want is the name for
each xml file I traverse ... I don't think I can do this with passing
parameters ...

-----Original Message-----
From: Jarno.Elovirta@xxxxxxxxx [mailto:Jarno.Elovirta@xxxxxxxxx]
Sent: Wednesday, August 28, 2002 9:57 AM
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
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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