Python: single instance of program on Windows using mutexes

Question

I create an executable using py2exe in python.
I was looking at this post but unfortunately the answers were superficial.
The first solution using Tendo but this works to limit 1 instance of application per user, and my app is being used in a Windows Server Environment where there are 20+ users loged in at a time.
And the second solution offered Listening to a defined port doesn't have examples of how it could be accomplished.
So I decided to go for mutexes to prevent my app from running multiple times.
So I currently use this code for using mutexes, but it doesn't have mutex detection between applications and Services.
This post shows how to accomplish mutexes, but doesn't show how its done in python.
How could I use Mutexes to have single instance of program on Windows where the mutexes don't limit single instance of program on Windows, and has detection between applications and Services.

Answer 1

I'm not sure why you'd have to use mutexes to this purpose on Windows? There's much simpler option: a bad old lockfile.

If all you want to achieve is making sure that only a single instance of the app runs, you could do something like this:

Windows supports you here since you can't delete a file if it's opened by another process. So (code untested):

tempdir = tempfile.gettempdir()
lockfile = os.sep.join([tempdir, 'myapp.lock'])
try:
    if os.path.isfile(lockfile):
        os.unlink(lockfile)
except WindowsError as e: # Should give you smth like 'WindowsError: [Error 32] The process cannot access the file because it is being used by another process..'   
    # there's instance already running
    sys.exit(0)

with open(lockfile, 'wb') as lockfileobj:
    # run your app's main here
    main()
os.unlink(lockfile)

with section ensures the file is opened when your main runs and it is closed when your main finishes running. Then os.unlink removes the lockfile.

If another instance tries to start up, it exits on WindowsError exception (it would be good to check its numeric code though to be sure it's precisely the case of the file already opened).

Above is a rough solution, a niftier one would be to use entry/exit functionality to delete lockfile if main exits for any reason. Explanation here: http://effbot.org/zone/python-with-statement.htm