[lxc-devel] [PATCH 2/3] lxc-ls: Show a simple error message when non-root

Serge Hallyn serge.hallyn at canonical.com
Tue Dec 4 21:34:49 UTC 2012


Quoting Stéphane Graber (stgraber at ubuntu.com):
> Instead of returning a python stacktrace, check what the current euid is
> and show an argparse error message similar to that used in lxc-start-ephemeral.
> 
> Signed-off-by: Stéphane Graber <stgraber at ubuntu.com>

Wait, you must be root to call lxc-ls at all?

> ---
>  src/lxc/lxc-ls | 7 +++++++
>  1 file changed, 7 insertions(+)
> 
> diff --git a/src/lxc/lxc-ls b/src/lxc/lxc-ls
> index 8a1d1ed..2ad1f7f 100644
> --- a/src/lxc/lxc-ls
> +++ b/src/lxc/lxc-ls
> @@ -32,6 +32,7 @@ warnings.filterwarnings("ignore", "The python-lxc API isn't yet stable")
>  import argparse
>  import gettext
>  import lxc
> +import os
>  import re
>  import sys
>  
> @@ -115,6 +116,12 @@ parser.add_argument("filter", metavar='FILTER', type=str, nargs="?",
>  
>  args = parser.parse_args()
>  
> +# Basic checks
> +## The user needs to be uid 0
> +if not os.geteuid() == 0:
> +    parser.error(_("You must be root to run this script. Try running: sudo %s"
> +                   % (sys.argv[0])))
> +
>  # --active is the same as --running --frozen
>  if args.active:
>      if not args.state:
> -- 
> 1.8.0
> 
> 
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