Date: Sat, 10 Nov 2007 07:45:43 +0200 From: Jonathan McKeown <jonathan+freebsd-questions@hst.org.za> To: freebsd-questions@freebsd.org Subject: Re: shell programming Message-ID: <200711100745.43287.jonathan%2Bfreebsd-questions@hst.org.za> In-Reply-To: <7EA6F842-7C0D-4EFC-BF4D-DD83B0524597@gmail.com> References: <47349D10.20709@ourweb.net> <7EA6F842-7C0D-4EFC-BF4D-DD83B0524597@gmail.com>
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On Friday 09 November 2007 20:02, Eric Crist wrote: > On Nov 9, 2007, at 11:46 AM, Bill Banks wrote: > > I'm writing a backup script. I need to get the day of the week into > > a variable. How can I do it? > > Well, it depends on what you're using. If you're using sh, see `man > date`. If you're using perl, it's quite complicated. Not really: use POSIX 'strftime'; my $day_of_week = strftime '%A', localtime; POSIX has always been a core module. To see this in action from a commandline, perl -MPOSIX=strftime -le 'print strftime q/%A/, localtime' Jonathan
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