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[xsl] obtaining name of output file

Subject: [xsl] obtaining name of output file
From: "Daniel Brauer" <daniel.brauer@xxxxxxxxxxxxx>
Date: Wed, 7 May 2003 16:50:32 +0200
Hi,

I need to obtain the name of the output file as specified at the command
line of the XSLT-processor.
Is there a way to do this?

Thanks
Daniel.




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