How do I keep a script running as a systemd service?

Question

I have a Python script with a function I want to execute every X minutes.

from threading import Timer

x = 5

def control():
    Timer(x*60, control).start()
    rest_of_the_function()

control()

When I run the script in its Python virtual environment from the terminal, there is no problem; when I run it as systemd service, it calls control() function just once and then it doesn't do anything else.

If I check the service through systemctl status myservice its Active field says active (running). There are no errors in the log.

The content of the /lib/systemd/system/myservice.service file is the following.

[Unit]
Description=Short description

[Service]
Type=simple
WorkingDirectory=/home/user/myservice
ExecStart=/home/user/myservice/myvenv/bin/python main.py

[Install]
WantedBy=multi-user.target

Why does it just call control() once? Can I fix it?

Answer 1

I think you are probably experiencing a simple buffering problem. If you are looking for messages on stdout, which by default is the journal for systemd units, then you may have to wait a long time for sufficient data to be printed before it gets flushed out.

A simple test for this is to add the -u option to python to make the output unbuffered. Remember to do systemctl daemon-reload after editing the service unit file.

You can also use strace to attach to the running python process. You will presumably find it waiting on a user futex, but after the appropriate time you should see the futex call return, and then entered again.

Answer 2

When using print() or anything else which writes to a file descriptor (file, pipe, etc), you need to flush the buffers with flush() if you need the output right away.

Alternatively for stdin, stdout and stderr you can use unbuffered mode using -u option for python described in the man page.