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From: | Vincent WATREMEZ |
Subject: | Re: Not understanding 'Program Status Testing' |
Date: | Tue, 22 Jul 2014 11:32:05 +0200 |
I have a daemon which I want to monitor specific status. I've created the following script called 'bucardo.monitor':
#!/bin/sh
bucardo status |grep mydb|cut -d"|" -f4| grep ".m" >/dev/null 2>&1
exit $?
In short, if the string "(one char)m" exists, I wish to get an alert. When I run the script from the command line, and the string I'm looking for exists, I get the following expected output:
me# bucardo.monitor;echo $?
0
I created a monit conf file thus:
alert address@hidden with reminder on 5 cycle
alert address@hidden with reminder on 5 cycle
check program bucardo-monitor with path /usr/local/bin/bucardo.monitor
with timeout 3 seconds
if status = 0 then alert
The manual states that the operator should be "==", however the last example under status only uses a single equals sign - and I've tried both, no difference. I've also use just "if status 0 then alert" as suggested in the manual, also no difference.
The problem is that monit always shows a last exit status of "1" - except for a few moments after issuing 'monit reload' to deploy changes to the script:
Program 'bucardo-monitor'
status Status ok
monitoring status Monitored
last started Mon, 21 Jul 2014 14:40:47
last exit value 1
data collected Mon, 21 Jul 2014 14:40:47
I've forced the test to be highly sensitive so that it will changed from an exit of 0 to 1 every few minutes, well within my monitoring window - but again, I never get a status other than 1 in monit status, and thus never get an alert.
Am I doing something wrong? Misunderstanding?
-- Paul Theodoropoulos www.anastrophe.com
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