How do I test (in one line) if command output contains a certain string?
BashBash Problem Overview
In one line of bash, how do I return an exit status of 0 when the output of /usr/local/bin/monit --version doesn't contain exactly 5.5 and an exit status of 1 when it does?
Bash Solutions
Solution 1 - Bash
! /usr/local/bin/monit --version | grep -q 5.5
(grep returns an exit-status of 0 if it finds a match, and 1 otherwise. The -q option, "quiet", tells it not to print any match it finds; in other words, it tells grep that the only thing you want is its return-value. The ! at the beginning inverts the exit-status of the whole pipeline.)
Edited to add: Alternatively, if you want to do this in "pure Bash" (rather than calling grep), you can write:
[[ $(/usr/local/bin/monit --version) != *5.5* ]]
([[...]] is explained in §3.2.4.2 "Conditional Constructs" of the Bash Reference Manual. *5.5* is just like in fileglobs: zero or more characters, plus 5.5, plus zero or more characters.)
Solution 2 - Bash
[ $(/usr/local/bin/monit --version) == "5.5" ]
eg-1: check for success
[ $(/usr/local/bin/monit --version) == "5.5" ] && echo "OK"
eg-2: check for failure
[ $(/usr/local/bin/monit --version) == "5.5" ] || echo "NOT OK"
or, to just check if the output contains 5.5:
[[ $(/usr/local/bin/monit --version) =~ "5.5" ]] || echo "NOT OK"
Solution 3 - Bash
Test the return value of grep:
sudo service xyz status | grep 'not' &> /dev/null
if [ $? == 0 ]; then
echo "whateveryouwant"
fi
I would recommend cron, it works fine with SALT stack