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 ________________________________________________________________________ This e-mail has been scanned for all viruses by Star Internet. The service is powered by MessageLabs. For more information on a proactive anti-virus service working around the clock, around the globe, visit: http://www.star.net.uk ________________________________________________________________________ XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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