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Date: Thu, 05 Nov 2009 09:26:15 -0400
From: PJ <af.gourmet@videotron.ca>
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To: Polytropon <freebsd@edvax.de>
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Cc: "freebsd-questions@freebsd.org" <freebsd-questions@freebsd.org>
Subject: Re: and now for conky & gremlins
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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?