The man page is very confusing. Yes, it says -c is class. But it also
has examples like this:
su -m operator -c 'shutdown -p now'
In my testing, this works:
$ su - root -c 'date'
Thu Aug 19 08:31:53 UTC 2021
and this does not:
$ su - root 'date'
date: No such file or directory.
What is -c supposed to do?
Ari
On 19/8/21 6:21pm, Andriy Gapon wrote:
On 2021-08-19 08:31, Aristedes Maniatis via freebsd-stable wrote:
I've got some scripts which are intended to run on a new EC2 instance
right after it is created. Since the script needs to install packages it
need to run as root. But because I don't have sudo installed at this
point (it is a brand new instance), I've only got 'su' to get root.
The script itself is launched over SSH with the ec2-user account and
there is no root password at this point in the startup.
My first attempt was to put this inside the script itself:
if ["$($whoami)" !="root" ];thenexec su -c"$0" exit1 fi
But su complains that I'm not allowed to execute a command using the -c
option as root.
-c option seems to be so confusing for some reason that it should bein
some FAQ document.
From the man page:
-c class
Use the settings of the specified login class. The login class
must be defined in login.conf(5). Only allowed for the super-
user.
You surely though that it did something else, right?
From the man page again:
If the optional args are provided on the command line, they are
passed to
the login shell of the target login. Note that all command line
arguments before the target login name are processed by su itself,
everything after the target login name gets passed to the login shell.
How else can I get this script running as root remotely in a completely
unattended way?
