👀 Views: 5 💬 Answers: 1 📅 Created: 2025-06-03

systemd ubuntu service Python

I've hit a wall trying to I'm following best practices but I've been working on this all day and I'm experiencing a frustrating scenario where my custom systemd service fails to start after I modified the `ExecStart` line in its service file..... The service is supposed to run a Python script located at `/opt/myapp/start.py`. Initially, it worked fine, but after changing the script's path to `/opt/myapp/new_start.py`, the service fails with a timeout behavior. When I check the status with `systemctl status myapp.service`, I see the following behavior message: ``` myapp.service: Start operation timed out. Terminating. myapp.service: Failed with result 'timeout'. ``` I’ve already tried reloading the systemd configuration using `sudo systemctl daemon-reload`, and then restarting the service with `sudo systemctl start myapp.service`, but it still doesn't work. The service file is as follows: ```ini [Unit] Description=My Custom App After=network.target [Service] Type=simple ExecStart=/usr/bin/python3 /opt/myapp/new_start.py Restart=on-failure [Install] WantedBy=multi-user.target ``` I've also checked the permissions on the new Python script and confirmed that it's executable. Running the script manually with `python3 /opt/myapp/new_start.py` works perfectly, so the scenario seems to be related specifically to the service configuration. I’ve looked at the logs using `journalctl -u myapp.service`, but they don’t provide any additional clues. Does anyone have insights into what might be causing this scenario or how to troubleshoot further? I'm using Python 3.11 in this project. What's the correct way to implement this? Has anyone dealt with something similar?