On 2005-11-03 16:33, Brandon Hinesley <[EMAIL PROTECTED]> 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. _______________________________________________ [email protected] mailing list http://lists.freebsd.org/mailman/listinfo/freebsd-questions To unsubscribe, send any mail to "[EMAIL PROTECTED]"
