Subject: RE: [xsl] obtaining name of output file From: "Michael Kay" <mhk@xxxxxxxxx> Date: Wed, 7 May 2003 20:56:45 +0100 |
> 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? > Pass it in as a parameter. (Write a simple shell script that uses the same value for the output file and the value of the xxxoutput-file stylesheet parameter.) Michael Kay XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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