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RE: Returning the file name of the input file

2002-08-28 04:01:01
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(_dot_)Elovirta(_at_)nokia(_dot_)com 
[mailto:Jarno(_dot_)Elovirta(_at_)nokia(_dot_)com]
Sent: Wednesday, August 28, 2002 9:57 AM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)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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