Linux distro name parsing

Question

Hello, I chose this way to get linux distro name:

ls /etc/*release

And now I have to parse it for name:

/etc/<name>-release

def checkDistro():
    p = Popen('ls /etc/*release' , shell = True, stdout = PIPE)
    distroRelease = p.stdout.read()

    distroName = re.search( ur"\/etc\/(.*)\-release", distroRelease).group()
    print distroName

But this prints the same string that is in distroRelease.

Answer 1

You need .group(1), because you want the first capture group - without arguments, it defaults to .group(0) which is the entire matched text.

Answer 2

Use .group(1).

Answer 3

What's the point? /etc/*release is not a standard, it will only work on some distros.

Answer 4

Parsing ls output is discouraged. Consider using a glob():

#!/usr/bin/env python

import os
import glob

def check_distro():
    print os.path.basename(glob.glob('/etc/*-release')[0]).replace('-release', '')

if __name__ == '__main__':
    check_distro()

Answer 5

An alternative is to use the builtin method platform.linux_distribution() (available in Python 2.6+):

>>> import platform
>>> platform.linux_distribution()
('Red Hat Enterprise Linux Server', '5.1', 'Tikanga')

In older versions of Python, platform.dist() can be used:

>>> import platform
>>> platform.dist()
('redhat', '5.1', 'Tikanga')