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Re: Program timed out

From: Russell Simpkins
Subject: Re: Program timed out
Date: Tue, 9 Jun 2015 08:29:53 -0400

Try putting that logic in a single bash script. With the && you're running two processes which creates two pids.



On Jun 9, 2015 8:13 AM, "Yuriy Zhilovets" <address@hidden> wrote:

For every program I am trying to run Monit responds: failed to start. Program timed out.
After a while it somehow notices that this program nevertheless runs and complains  about this sad fact.
No difference if I check process by its PID file of with a "match" directive

Monit 5.11 under Ubuntu 15.04


check process move-data matching ""
  start program = "/bin/bash -c 'cd /var/opt/poller && ./'"
  stop program = "/usr/bin/pkill -f"
================================== just enters an infinite loop and never returns. No forks, detaching from terminal or something of this kind

$ monit -Ivv

'move-data' start: /bin/bash
'move-data' failed to start (exit status -1) -- Program /bin/bash timed out
    monit() [0x41cf7b]
    monit(LogError+0xb6) [0x41d906]
    monit(Event_post+0x291) [0x4197c1]
    monit() [0x417fc4]
    monit(control_service+0x137) [0x418897]
    monit() [0x42b6bd]
    monit(validate+0x78) [0x42cc88]
    monit(main+0x482) [0x40c4a2]
    /lib/x86_64-linux-gnu/ [0x7effdf088a40]
    monit() [0x40c771]
'move-data' monitoring enabled
'move-data' start action done
'move-data' check skipped -- service already handled in a dependency chain
'move-data' process is running with pid 14570
'move-data' process is running after previous exec error (slow starting or manually recovered?)
'move-data' zombie check succeeded [status_flag=0000]
'move-data' process is running with pid 14570

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