The pythonic way equal to "whoami"

From	TheSaint <nobody@nowhere.net.no>
Newsgroups	comp.lang.python
Subject	The pythonic way equal to "whoami"
Followup-To	comp.lang.python
Date	2011-06-07 22:24 +0800
Organization	Aioe.org NNTP Server
Message-ID	<islca0$asr$1@speranza.aioe.org> (permalink)

Followups directed to: comp.lang.python

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Hello,
I was trying to find out whose the program launcher, but os.environ['USER'] 
returns the user whom owns the desktop environment, regardless the program 
is called by root.
I'd like to know it, so the program will run with the right privileges.

Is there any standard function on python, that will do it?
-- 
goto /dev/null

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The pythonic way equal to "whoami" TheSaint <nobody@nowhere.net.no> - 2011-06-07 22:24 +0800
  Re: The pythonic way equal to "whoami" Kushal Kumaran <kushal.kumaran+python@gmail.com> - 2011-06-07 20:29 +0530
    Re: The pythonic way equal to "whoami" TheSaint <nobody@nowhere.net.no> - 2011-06-08 21:58 +0800
      Re: The pythonic way equal to "whoami" Christopher Head <chead@is.invalid> - 2011-06-08 13:08 -0700
        Re: The pythonic way equal to "whoami" TheSaint <nobody@nowhere.net.no> - 2011-06-09 19:34 +0800
      Re: The pythonic way equal to "whoami" Nobody <nobody@nowhere.com> - 2011-06-09 03:38 +0100

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