Hello there,
I am trying to run an python script upon boot-up
I followed the instructions from “michelangelo” step by step.
After I created the the file myscript.service I can run the Code:
“systmectl enable myscript.service” w/o error message
With command “systemctl status myscript.service”
I get the following output:
Sep 26 18:46:36 volumio-2 systemd[1]: myscript.service: main process exited, code=exited, status=203/EXEC
Sep 26 18:46:36 volumio-2 systemd[1]: Unit myscript.service entered failed state.
Sep 26 18:46:36 volumio-2 systemd[1]: myscript.service holdoff time over, scheduling restart.
Sep 26 18:46:36 volumio-2 systemd[1]: Stopping Volumio Backend Module…
Sep 26 18:46:36 volumio-2 systemd[1]: Starting Volumio Backend Module…
Sep 26 18:46:36 volumio-2 systemd[1]: myscript.service start request repeated too quickly, refusing to start.
Sep 26 18:46:36 volumio-2 systemd[1]: Failed to start Volumio Backend Module.
Sep 26 18:46:36 volumio-2 systemd[1]: Unit myscript.service entered failed state.
I have tried to use a very simple python script with only the shebang and a “pass” in it to make sure the script is not the issue.
I have also used the “systemctl start myscript.service” even so it is not mentioned in the thread.
Question(s):
- What can be done to avoid this issue?
- Is there another way to start a python script with root privileges?
- Furthermore I need to run my script with python3
- Is there a typo:
======= Quote ======================================================
Save it, then enable it
CODE:
systemctl enable myscript.service
then start it with
CODE:
systemctl status myscript.service
and finally, check if its running with
CODE:
systemctl status myscript.service
Thank you very much in advance.
Any hint is welcomed