[xsl] Returning the file name of the input file

[Cams Ismael <Ismael.Cams@xxxxxxxxxxxxxxx>]

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.
I am afraid there is no method to do this, but if anybody knows a solution I
will be very thankfull for that.


kind regards,
Ismaël





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