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Re: [xsl] obtaining name of output file

Subject: Re: [xsl] obtaining name of output file
From: David Carlisle <davidc@xxxxxxxxx>
Date: Wed, 7 May 2003 16:26:13 +0100
  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?

Declare it as a global parameter of the stylesheet and set this parameter
on the command line at the same time as setting the output destination.

David

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