python：运行'nohup'：如何摆脱消息？

Question

I have a variable list of programs that I want to kick off from a cron job. The solution I have settled on at least for now is to write the actual cron job in python, then run through the list, starting each program with:

outf=open('the_command.log','w')
subprocess.Popen(['nohup','the_command', ...],stdout=outf)
outf.close()

The problem with this is that it creates a nohup.out file - the same one for each process, it seems. If I did this same thing from the command line, it might look like:

$ nohup the_command ... > the_command.log 2>&1 

This works fine, except I get a message from nohup when I run it:

nohup: ignoring input and redirecting stderr to stdout

I have tried to redirect stderr to /dev/null, but the result is that the_command.log is empty. How can I solve this?

Answer 1

I solved this by using a different command detach from http://inglorion.net/software/detach/

But I now consider this to be improper. It would be better to use oneshot services started by your cron job script or make your cron entry cause a oneshot service to be started.

With this there would be no need to detach as the processes aren't your scripts children rather they are children of the supervisor. Any init that supports starting a normally down service and does not restart it when it exits can be used.