Continue while loop in Bash Script even if a condition fails - bash

Question

#!/bin/bash
var="true"
i=1
while $var
do
read -p "Enter value (true/false): " var
if [[ $var == "true" ]]
then
echo "Iteration : $i"
((i++))
elif [[ $var == "false" ]]
then
echo "Exiting the process"
elif [[ $? -eq 1 ]]
then
echo "Invalid Choice."
echo "Avaialable Choices are true or false"
exit
fi
done
Script is Working Fine. I Enter true the loop will iterate for false the script stops.
I want the script will continue asking "Enter Value" if any other value instead of true or false will be entered.

Answer 1

This would do the same with a more academic syntax:
i=0
while :; do
printf 'Enter value (true/false): '
read -r var
case $var in
true)
i=$((i + 1))
printf 'Iteration : %d\n' $i
;;
false)
printf 'Exiting the process\n'
break
;;
*)
printf 'Invalid Choice.\nAvaialable Choices are true or false\n'
;;
esac
done

Answer 2

You might find this to be a cleaner solution:
i=0
while true; do
read -p "enter value: " myinput
if [[ $myinput = true ]]; then
echo "iteration $i"
i=$((i+1))
elif [[ $myinput = false ]]; then
echo "exiting"
exit
else
echo "invalid input"
fi;
done;
The issue I see with your current code is that it is unclear which command's exit status $? refers to. Does it refer to the echo in the previous elif block? Or the last condition check? Or something else entirely?

Answer 3

I'm new in bash. I tried that:
#!/bin/bash
i=1
while [[ $var != "false" ]]
do
read -p "Enter value (true/false): " var
if [[ $var == "true" ]]
then
echo "Iteration : $i"
((i++))
elif [[ $var == "false" ]]
then
echo "Exiting the process"
elif [[ $? -eq 1 ]]
then
echo "Invalid Choice."
echo "Avaialable Choices are true or false"
fi
done
I changed while $var with while [[ $var ]] because while works like if. It runs the given command. In there it is $var's value.
And I moved exit to first elif expression's end. So if user type false program will exit.