Date: Thu, 05 Nov 2009 09:26:15 -0400 From: PJ <af.gourmet@videotron.ca> To: Polytropon <freebsd@edvax.de> Cc: "freebsd-questions@freebsd.org" <freebsd-questions@freebsd.org> Subject: Re: and now for conky & gremlins Message-ID: <4AF2D277.3090406@videotron.ca> In-Reply-To: <20091105023045.9a3d90ab.freebsd@edvax.de> References: <4AF1FF76.60808@videotron.ca> <20091105023045.9a3d90ab.freebsd@edvax.de>
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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?
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