Not understanding 'Program Status Testing'

[Paul Theodoropoulos]

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