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Date: Fri, 4 Nov 2005 03:29:22 +0200
From: Giorgos Keramidas <keramida@ceid.upatras.gr>
To: Brandon Hinesley <brandonh@hotandcold.biz>
Message-ID: <20051104012922.GA1025@flame.pc>
References: <20051103215644.GA34288@flame.pc>
	<002401c5e0d7$64f3b1d0$6800a8c0@BrandonH>
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In-Reply-To: <002401c5e0d7$64f3b1d0$6800a8c0@BrandonH>
Cc: freebsd-questions@freebsd.org
Subject: Re: Cron Job will not run.
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On 2005-11-03 16:33, Brandon Hinesley <brandonh@hotandcold.biz> wrote:
>
> Okay, the problem seems to be with a certain part of my script.
> Like I said, it works fine when I start it manually (./)
>
> I set up a few "checkpoints" if you will, and I determined that
> it's this loop that cron has a problem with:
>
> for (( i = $numbkups ; i >= 2 ; i-- ))
> do
>     let from=i-1
>     mv -fv $dbkups/$from $dbkups/$i
> done
>
> This loop never runs and neither does anything after it.  I
> don't see why cron would have any problem with this, or why
> this would exit the script...

Hmmm, what shell is this supposed to run in?
It doesn't look like /bin/sh syntax to me.