Date: Thu, 5 Nov 2009 15:46:09 +0100 From: Ruben de Groot <mail25@bzerk.org> To: PJ <af.gourmet@videotron.ca> Cc: Polytropon <freebsd@edvax.de>, "freebsd-questions@freebsd.org" <freebsd-questions@freebsd.org> Subject: Re: and now for conky & gremlins Message-ID: <20091105144609.GA28950@ei.bzerk.org> In-Reply-To: <4AF2D277.3090406@videotron.ca> References: <4AF1FF76.60808@videotron.ca> <20091105023045.9a3d90ab.freebsd@edvax.de> <4AF2D277.3090406@videotron.ca>
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On Thu, Nov 05, 2009 at 09:26:15AM -0400, PJ typed: > Polytropon wrote: > > On Wed, 04 Nov 2009 18:25:58 -0400, PJ <af.gourmet@videotron.ca> wrote: > > > >> output should be: 1 2 3 [4] 5 6 7 etc. > >> is: 1 2 3 4 5 6.... > >> > >> the calendar.sh is exactly: > >> #!/bin/sh > >> cal | awk 'NR>1' | sed -e 's/ / /g' -e 's/[^ ] /& /g' -e 's/..*/ > >> &/' -e "s/\ `date +%d`/\[`date +%d`\]/" > >> > > > > It's quite obviously. Let's try the last substitution > > argument in plain shell: > > > > % date +%d > > 05 > > > > But the command creates this: > > > > Su Mo Tu We Th Fr Sa > > 1 2 3 4 5 6 7 > > > > The leading zero is missing, so there's no substition that > > changes "5" into "[5]", because the search pattern is "05". > > > Ok, I see... I'm not too good in programming. I guess I didn't notice > the previous to the first days of November the date was always 2 > digits.. how do I get rid of the zero? Regex substitution or something > like that? date "+%e" should do it. Ruben
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