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Date: Thu, 5 Nov 2009 15:46:09 +0100
From: Ruben de Groot <mail25@bzerk.org>
To: PJ <af.gourmet@videotron.ca>
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	PJ <af.gourmet@videotron.ca>, Polytropon <freebsd@edvax.de>,
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Cc: Polytropon <freebsd@edvax.de>,
	"freebsd-questions@freebsd.org" <freebsd-questions@freebsd.org>
Subject: Re: and now for conky & gremlins
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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