Systemd Daemon in Python with watchdog support

Question

I want to write a daemon in python which gets started via systemd.

I want to use Type=notify, this way I don't have to do the double fork magic.

According to the docs:

The reference implementation for this notification is provided by libsystemd-daemon.so

... how to do this with Python?

Horner

Possibly related: freedesktop.org/software/systemd/python-systemd/daemon.html #13070134 – Protection

@Robᵩ The question you mention does not solve my question, since my question is about a daemon with watchdog support. If my daemon hangs in an endless loop, I want systemd to know that the service does not respond. Without watchdog systemd things my service is alive as long as the process runs. – Horner

Answer 1

Probably, you could use sdnotify python module which is a pure-python implementation of sd_notify protocol. Actually, the protocol is rather simple, so the module implementation is quite short.

To use watchdog machinery you should add WatchdocSec=<smth> to the unit file, and then send WATCHDOG=1 messages on a regular basis from your service. Check Restart= option as well.

Selfreliant

Nice, the package can be installed via pip. Thank you. – Horner

It's how I've discovered it a while ago :). pip search facility is rather handy. – Selfreliant

Answer 2

use the package
https://pypi.org/project/systemd-python/
it is the offical systemd devs and maintained.

Pillion

You didn't answer the question. Give an example of how to notify systemd from python using that package. – Dade

Answer 3

You can use the systemd-python package. Specifically, consider the notify function in the daemon module.

from systemd.daemon import notify
notify("WATCHDOG=1")

According to the docs this will...

Send a message to the init system about a status change.

Centralism

Answer 4

I implemented the functionality in systemd-watchdog-thread

Githens