while loop on bash scripting

Question

My code is :

#!/bin/bash

while true; do

    COUNTER=$((COUNTER +1))     
    echo $COUNTER

    if (($COUNTER%3 == 0));  then
        echo "Counter now is 3"
        sleep 2
    fi

    if (($COUNTER%6 == 0));  then
        echo "Counter now is 6"
        sleep 2
    fi

    if (($COUNTER%9 == 0));  then
        echo "Counter now is 9"
        sleep 2
    fi

    if (($COUNTER%12 == 0));  then
        echo "Counter now is 12"
        sleep 2
    fi

    if (($COUNTER%15 == 0));  then
        echo "Counter now is 15"
        sleep 2
        exit
    fi

done

and my output is :

./test2.sh 
1
2
3
Counter now is 3
4
5
6
Counter now is 3
Counter now is 6
7
8
9
Counter now is 3
Counter now is 9
10
11
12
Counter now is 3
Counter now is 6
Counter now is 12
13
14
15
Counter now is 3
Counter now is 15

Why is it showing, every time, Counter now is 3 or sometime show more on echo. I don't want it to show additional on echo

Answer 1

It would be more correct if your outputs said what the corresponding test tested for, e.g.

if (( COUNTER%3 == 0 )); then
    echo 'COUNTER is a multiple of 3'
fi

This is the test that you have, and if you think about it, 6, 9, 12, and 15 are also multiples of 3.

In this short script, I would suggest that you test directly on COUNTER rather than on COUNTER%somevalue if you want to check the actual value of the counter rather than whether it's a multiple of some value:

if (( COUNTER == 3 )); then
    echo 'COUNTER is 3'
fi

Note that $ is not needed on variables when they are used in an arithmetic context.

Answer 2

Because output is a lie.

• What is output when it says “count is…”?
• What would be a better text than “Count is”?
• Where does it echo count? (make it echo count is …)

When you have fixed that, you should understand what it is doing.