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Date:      Sat, 17 May 2003 21:44:28 +0200
From:      =?ISO-8859-1?Q?Rapha=EBl_Marmier?= <raphael@computer-rental.ch>
To:        Niklas Saers Mailinglistaccount <niklasmls@doriath.saers.com>
Cc:        FreeBSD-Questions@freebsd.org
Subject:   Re: bash shell for-loop
Message-ID:  <F8AA746A-889F-11D7-A528-000393D67E4A@computer-rental.ch>
In-Reply-To: <20030517164813.V37832@doriath.saers.com>

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Looks like it stop to ask fo the password...

Raphael

Le samedi, 17 mai 2003, =E0 16:55 Europe/Zurich, Niklas Saers=20
Mailinglistaccount a =E9crit :

> Hi,
> I've created a small bash script that stops, and I cannot understand=20=

> why
> it stops. It is:
>
> #!/usr/local/bin/bash
> for username in $(ls /home); do
>   su -m $username /home/$username/startup.sh;
> done
>
> When I type this at the bash prompt, it goes well. However, when I =
call
> this script it executes the first su and then gives me:
>
> [4]+  Stopped                 bash /tmp/sh.sh
>
> Why does it stop? I've nowhere asked it to stop. I certainly don't=20
> want it
> to stop.
>
> To explain my command, I start a set of daemons for each user=20
> specified in
> ~/startup.sh and no user has a valid shell as I do not wish to allow=20=

> them
> any kind of login access. Thus I have to include the "-m"
>
> What makes a bash script stop? How can I set it to ignore whatever is
> requesting it (errorlevels?) to stop?
>
> Cheers
>
>   Nik
> _______________________________________________
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> http://lists.freebsd.org/mailman/listinfo/freebsd-questions
> To unsubscribe, send any mail to=20
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