Date: Thu, 5 Nov 2009 02:30:45 +0100 From: Polytropon <freebsd@edvax.de> To: PJ <af.gourmet@videotron.ca> Cc: "freebsd-questions@freebsd.org" <freebsd-questions@freebsd.org> Subject: Re: and now for conky & gremlins Message-ID: <20091105023045.9a3d90ab.freebsd@edvax.de> In-Reply-To: <4AF1FF76.60808@videotron.ca> References: <4AF1FF76.60808@videotron.ca>
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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". -- Polytropon Magdeburg, Germany Happy FreeBSD user since 4.0 Andra moi ennepe, Mousa, ...
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