How to call another script within a script with bash - bash

Question

Hello here is my primary script. The test2.sh is just an echo "it worked"
what happens when I try and call from the original loop, it gets to the correct file then echo's infinite "it worked" where it should just be doing it once.
Any idea why? I really want to have another loop called outside of the main script that won't interfere, but still learning bash =P
#!/bin/bash
number=1
while true
do
if [ "$number" -eq "1" ]; then
echo "hello 1!"
elif [ "$number" -eq "2" ]; then
echo "hello 2!"
elif [ "$number" -eq "3" ]; then
echo "hello 3!"
elif [ "$number" -eq "4" ]; then
./test2.sh & continue
fi
sleep 5
((number++))
echo $number
done

Answer 1

first observation & is not a logical operator, & runs the precedding command in the background. Use && for logical operations.
what you need is a break keyword not a continue keyword . If you use the break keyword, the loop will stop executing. The continue keyword only rexecutes the loop , and since number is 4 , this branch of code will always run elif [ "$number" -eq "4" ]; then
working code
#!/bin/bash
number=1
while true
do
if [ "$number" -eq "1" ]; then
echo "hello 1!"
elif [ "$number" -eq "2" ]; then
echo "hello 2!"
elif [ "$number" -eq "3" ]; then
echo "hello 3!"
elif [ "$number" -eq "4" ]; then
./test2.sh && break
fi
sleep 5
((number++))
echo $number
done
or you can do this
for number in {1..4};do
(( number == 4 )) && ./test2.sh || echo "$number"
sleep 5
done