I've recently began using ConfigParser() for my python scripts to add some functionality to them for config files. I know how to use it but I have a problem. My script needs to run as the root user, using sudo. The config files are in ~/.config/scriptconfig/
but when you run a script as sudo it temporarily changes users to root, so it doesn't find the config files. What I want to do is get the config file of the effective user so it grabs /home/myuser/.config/scriptconfig/config.cfg
instead of /root/.config/scriptconfig/config.cfg
, which doesn't exist.
I need the script to be able to run on different machines, not just mine. So I need to get the home directory of the effective user
Here is an example of the code I'm trying to use:
import os, ConfigParser
config = ConfigParser.RawConfigParser()
homepath = os.path.expanduser("~/")
configpath = homepath + ".config/scriptconfig/config.cfg"
config.read(configpath)
get = config.get('Config', 'Example')
print get
It should print the value of example
from the config file but when ran as sudo, the path is /home/root
so it doesn't find the config file.
If you run your script with sudo (sudo myscript.py
) then the environment variable $USER
will be root
and the environment variable $SUDO_USER
will be the name of the user who executed the command sudo myscript.py
. This following is simply a clarification of the previous post by Cédric Julien. Consider the following scenario:
A linux user bob
is logged into the system and possesses sudo privileges. He writes the following python script named myscript.py
:
#!/usr/bin/python
import os
print os.getenv("USER")
print os.getenv("SUDO_USER")
He then makes the script executable with chmod +x myscript.py
and then executes his script with sudo privileges with the command:
sudo ./myscript.py
The output of that program will be (using python 2.x.x):
root
bob
If bob runs the program sans sudo privileges with
./myscript.py
he will get the following output:
bob
None