Hello I'm writing a daemon in python which uses the python-daemon module, my application starts correctly, there is a pidfile.lock created but no sign of the pidfile containing the process id.
import daemon
import lockfile
import perfagentmain
context = daemon.DaemonContext(
working_directory='/opt/lib/perf-agent',
umask=0o002,
pidfile=lockfile.FileLock('/var/run/perf-agent.pid')
)
with context:
perfagentmain.start()
I agree with @npoektop 's comment about the solution. I would just say that daemon.pidlockfile
does not exist at the time I am writing this. daemon.pidfile
instead. Maybe that's a recent name change?
So instead, here's the general solution using the daemon.pidfile
module instead of the lockfile
module.
import daemon
import daemon.pidfile
import perfagentmain
context = daemon.DaemonContext(
working_directory='/opt/lib/perf-agent',
umask=0o002,
pidfile=daemon.pidfile.PIDLockFile('/var/run/perf-agent.pid')
)
with context:
perfagentmain.start()
And @Martino Dino, you're absolutely right, it seems the lockfile
module has a totally different implementation of writing lock files. (even though python-daemon
actually requires lockfile
)
When I tried out pidfile = lockfile.FileLock('/var/run/mydaemon.pid')
for my own needs, I instead saw a file called <MY_MACHINE_NAME>-<8CHAR_HEX_ID>.<PID_OFF_BY_2>
, along with a file /var/run/mydaemon.pid.lock
. This answer mentions how this method of hard linking a randomly named file to your pidlock file was a file-locking method prior to the use of the O_EXCL
flag used when opening files.
But the annoying part was that the file did not contain the PID as you said, and the file name had a PID which was off by a few numbers of the correct PID, so it was terribly misleading.
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