systemd service start on boot

README.md

README

Goal

Have a command run at boot

Summary

In 2024, we use systemd for this in all major distributions. There are other distributions that do not like systemd and are still using older initialization ("init") systems like "Sys-V" or alternatives like "runit". If you are unsure, you're probably running systemd and should follow this guide. You would usually know if you're using an alternative.

Requirements

• Debian-based OS (Debian, Ubuntu, Mint, Raspbian)
• RedHat-based OS (RHEL, CentOS, Rocky, Alma, Suse)

Instructions

1. Pick a service name, like "script" or "server". We will use NAME to be a placeholder but you should swap in your service name
2. Create or edit a file at /etc/systemd/system/NAME.service (See contents below)
3. Add more environment variables, edit the parameters as needed.
4. You may execute a script file (make sure it's executable!) or a command like you would on the command line
5. If you only want the script to run once, instead of automatically restarting, switch "simple" to "oneshot"
6. Run sudo chmod 660 /etc/systemd/system/NAME.service
7. Run sudo systemctl daemon-reload to re-scan the files (we need to do this because you just created/edited one)
8. Run sudo systemctl start NAME
9. Verify your service started correctly with systemctl status NAME
10. View logs if needed with sudo journalctl -n 20 -u NAME (returns the latest 40 lines)
11. If you need to, you can restart the service with sudo systemctl restart NAME
12. Run systemctl enable NAME to start this service automatically on boot/startup
13. Test it by rebooting your system now

NAME.service

[Unit]
Description=myservice

[Service]
Type=simple
User=user
Group=user
Environment=SOMEVAR=myvalue
WorkingDirectory=/home/user
ExecStart=/home/user/scripts/script.sh
Restart=always
RestartSec=3
StartLimitBurst=5
StartLimitInterval=0
StandardOutput=journal
StandardError=journal+console
SyslogIdentifier=myservice
TimeoutSec=60

[Install]
WantedBy=multi-user.target