Shell script + time dependency - linux

Question

I wanted to write a shell script for the following:
I want to check if a service is running, if it is running then exit 1, or else after 5 min of it not running, exit -1.
something like:
while(for 5 minutes) {
if service running, exit 1
}
exit -1 //service is not running even after 5 minutes, so exit -1.
I am able to check the condition that if service is running or not, but not able to add the time constraint part.
This is what i attempted
if (( $(ps -ef | grep -v grep | grep tomcat7 | wc -l) > 0 ));
then
echo "running"
else
echo "NOT running"
fi

Answer 1

You should use the bash sleep command. An excerpt from the man page:-
You could provide sleep 5m in your script to wait for 5 minutes and do an action.
NAME
sleep - delay for a specified amount of time
DESCRIPTION
Pause for NUMBER seconds. SUFFIX may be 's' for seconds (the default), 'm' for minutes, 'h' for hours or 'd' for days. Unlike most implementations that require NUMBER be an
integer, here NUMBER may be an arbitrary floating point number. Given two or more arguments, pause for the amount of time specified by the sum of their values.
A proper way to your solution would be:-
#!/bin/bash
maxAttempts=0
maxCounter=2 # Number of attempts can be controlled by this variable
while [ "$maxAttempts" -lt "$maxCounter" ]; do
if ps -ef | grep -v grep | grep "tomcat7" > /dev/null
then
echo "tomcat7 service running, "
exit 1
else
maxAttempts=$((maxAttempts+1))
sleep 5m # The script waits for 5 minutes before exiting with error code '-1'
fi
done
exit -1
The if condition works when the ps -ef | grep -v grep | grep "tomcat7" returns a command success error code for which the condition passes. > /dev/null to suppress all standard outpur(stdout, stderr) to /dev/null so that we can work only with the exit codes of the command provided.