[Top][All Lists]

[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]

Re: check program reports exit codes that I can't recreate

From: Jan-Henrik Haukeland
Subject: Re: check program reports exit codes that I can't recreate
Date: Fri, 31 Oct 2014 21:22:54 +0100

If you remove the “with timeout 3 seconds” or increase the timeout value the 
“problem" should go away. If the timeout statement is used, Monit will kill 
your program if it uses more than _timeout_ time to run. This seems to be the 
case here and the exit value 9 comes from Monit sending the kill signal to the 
process (by convention).

> On 31 Oct 2014, at 18:15, Russell Simpkins <address@hidden> wrote:
> I have a check:
> check program foo with path "/usr/bin/ 7005" with timeout 3 seconds
>   if start != 0 then
>   start with program "/etc/init.d/myservice" with timeout 10 seconds
> My script will either exit with a zero or a one. Monit runs as root. I'm 
> running monit 5.8 Sometimes monit reports error codes that I can't recreate 
> e.g.
> Oct 15 13:56:36 ip-10-45-179-177 monit[655]: 'foo' '' failed with 
> exit status (9) -- no output from program
> I can't figure out why monit gets an exit status that I can't recreate. I've 
> changed my check to be
> if start == 1 then
> That seems to have fixed my problem. I also tried the following in my bash 
> script:
> set -e
> Thinking that maybe one of my steps was failing, but I can never recreate any 
> other exit code besides 0 when running it by hand. Is this something fixed in 
> 5.9? 

reply via email to

[Prev in Thread] Current Thread [Next in Thread]