CODEWITHSUNDEEP
pythondjangosubprocesssupervisord
lukik
lukik

Reputation: 4070

How do I restart a python app from within the app itself

Got a Django app that has users going through a setup wizard. After the wizard, the app needs to restart for it to pick the necessary settings.

The app is run on Ubuntu 18.04 and monitored using supervisord.

Ideally am looking to call systemctl restart supervisord.service from the Django app itself.

So I've tried

import subprocess
subprocess.run("systemctl restart supervisord.service")

However, this fails with the error:

FileNotFoundError [Errno 2] No such file or directory: 'systemctl restart supervisord.service': 'systemctl restart supervisord.service'

There is this question here on SO but that is an older question and the answers there are relying on os.* while as of this posting subprocess seems to be the preferred way or accessing OS function

Upvotes: 1

Views: 1189

Answers (2)

Rahul Rao
Rahul Rao

Reputation: 16

For more cleaner way to start your own python program would be:- 

import os
import sys
import psutil
import logging

def restart_program():
    """Restarts the current program, with file objects and descriptors
       cleanup
    """

    try:
        p = psutil.Process(os.getpid())
        for handler in p.get_open_files() + p.connections():
            os.close(handler.fd)
    except Exception, e:
        logging.error(e)

    python = sys.executable
    os.execl(python, python, *sys.argv)

Upvotes: 0

lukik
lukik

Reputation: 4070

Seen my error. The command should be run as:

subprocess.run(["systemctl", "restart", "supervisor.service"])

source: http://queirozf.com/entries/python-3-subprocess-examples

Upvotes: 1

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