Date: Fri, 4 Nov 2005 03:29:22 +0200 From: Giorgos Keramidas <keramida@ceid.upatras.gr> To: Brandon Hinesley <brandonh@hotandcold.biz> Cc: freebsd-questions@freebsd.org Subject: Re: Cron Job will not run. Message-ID: <20051104012922.GA1025@flame.pc> In-Reply-To: <002401c5e0d7$64f3b1d0$6800a8c0@BrandonH> References: <20051103215644.GA34288@flame.pc> <002401c5e0d7$64f3b1d0$6800a8c0@BrandonH>
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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.
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