How to return different exit code for if and else in bash?

Shell Programming and Scripting
1 
#!/bin/sh
db_time=$(mysql -u root -BNe 'USE monit_sikdai; SELECT MAX(alert_date) FROM alerts')
echo $db_time
# 2024-05-31
system_time=$(date +"%Y-%m-%d %H:%M:%S")
db_timestamp=$(date -d "$db_time" +%s)
system_timestamp=$(date -d "$system_time" +%s)
difference=$((system_timestamp - db_timestamp))
if [ $difference -gt 600 ]; then
    echo "alert"
else
    echo "do nothing"
fi

I want to take this value into monit like this.

 check program myscript with path /root/check_db.sh
       if status != 0 then alert

So, I want to send alert when status not equals to some value(say 0).

So, I want if to generate exit status of not zero whereas else to generate exit status of 0. So, I don't get alert when else is executed whereas, I get alert when if gets executed.

I would also want to wait 10 times till I get the differences > 600 before sending alert.

2

Not sure I understood...
What's wrong with in your if cond. instead of (echo "alert" ) you had directly ( exit 20 ) or any other value you wished...

3

Since this is localhost, I am just trying to dry run how the alert will work. You're correct.

4

You wrote in your title returning different exit code...
An exit code is different than an alert, an alert is a message you put, that you check in the output... and exit code is a stop of execution with a code value for you to know where in your script it stopped

5

loop up to 10 times with some sort of a delay between iterations,

example:

#!/bin/bash

MAXSLEEP=15
MAXRETRIES=10
iteration=1

while [ $iteration -le $MAXRETRIES ]
do
        echo 'existing code .... '

        echo "iteration $iteration of $MAXRETRIES"

        sleep $((1 + $RANDOM % $MAXSLEEP))
        ((iteration++))
done

this is trivial, you should be more than capable of these type of activities given your declared level of experience.

6

I mean by monit there are cycles or something in monit. Tried to understand how they work.

7

I've designed the program in such a way that it returns exit code 1 when if is true, which I'll check with monit.


check program myscript with path /root/check_db.sh
       if status != 0 then then exec "send-alert.sh"

Now, I want to execute this only if status!=0 for 10 cycles.

8

To be clear: the child process returns an un-named status value with exit.

Its parent process must collect that value immediately, with a command like:

status="${?}"

and you would check it like:

if (( status != 0 )); then send-alert.sh; fi

The reason for collecting the status immediately is that almost all Bash commands set a new status. Even an echo of the value as debug, or the first of multiple tests, will change it.

$ 
$ bash  ## Open a child shell.
<my subshell> $ exit 27  ## Exit with status
exit
$ echo "${?}"
27
$ echo "${?}"
0
$

The second echo reports the success of the first echo, not the status of the previous shell as might be expected.

9

I didn't want to ask a new thread. I wrote this script. But the problem is it queries database every 1 monit cycle. Since the target Alerts don;t go at night as no transactions occur at night, it'll keep alerting and spamming emails.
Is there a clever way to check if a system is functioning correctly or not?

The server is a java web server consisting of wrapper.log. I tried to telnet its ports but I suspect the ports of this server change at restart(Not sure).

Any guidance? I need to find if a module in linux server is up or down.

First I need to define the state UP. I am not sure how do I do it.

10

There must be a function in monit to withhold/delay notifications...