Hello, I want to monitor a PHP script but I get a Sytax error .
This is the monit script:
check program test with path "/home/user/test.php"
if status != 0 then alert
I always get
/etc/monit/monitrc:23: Error: syntax error 'test'
test.php is executable and contains :
#!/usr/bin/php -q
<?php
$err=1 ;
if($err != 0) exit(1);
?>
If I run it in shell I get the correct $? of 0 or 1
Anyone knows how to solve this?
Thank you
P.
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