Systemd Daemon in Python with watchdog support

Question

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

I wan 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?

Possibly related: https://www.freedesktop.org/software/systemd/python-systemd/daemon.html http://stackoverflow.com/questions/13069634/python-daemon-and-systemd-service — Robᵩ, Apr 27 '16 at 15:22
@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. — guettli, May 02 '16 at 11:38

user3159253 · Accepted Answer · 2016-05-09T09:07:02.653

9

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.

edited May 09 '16 at 09:07

answered May 09 '16 at 08:57

user3159253

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Nice, the package can be installed via pip. Thank you. – guettli May 09 '16 at 10:33
It's how I've discovered it a while ago :). `pip search` facility is rather handy. – user3159253 May 09 '16 at 10:36

score -1 · Answer 2 · answered Apr 25 '20 at 11:55

-1

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

answered Apr 25 '20 at 11:55

qrtLs

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You didn't answer the question. Give an example of how to notify systemd from python using that package. – shrewmouse Sep 23 '20 at 20:48

Systemd Daemon in Python with watchdog support

2 Answers2