Bash Script verify user input is not empty and is equal to a value

Shell Programming and Scripting
1

I need to create a script that has a user enter a value. I want to verify that the value is either 1,2, or 3. If it is not then I want them to try entering it again. I am using a while loop to force them to retry.

I am able to test the input against 1,2, and 3, but when I test agains an empty variable, I get an error (expected unary). Below is what I have so far:

echo
echo "Which item?"
echo "1) choice 1"
echo "2) choice 2"
echo "3) choice 3"
echo
echo "Enter Choice:"
read -e INPUT
echo
while [ $INPUT != 1 ] && [ $INPUT != 2 ] && [ $INPUT != 3 ] && [ $COUNTER -lt 5 ] || [ -z $INPUT ]; do
        clear
        echo
        echo "You did not enter an appropriate choice.  Please enter 1, 2, or 3."
        echo "1) choice 1"
        echo "2) choice 2"
        echo "3) choice 3"
        echo
        echo "Enter Choice:"
        read -e INPUT
        echo
        let COUNTER++
        echo "Counter =$COUNTER"
done
# PROVIDER VARIABLES
case $INPUT in
1)
        PROV_POOL="choice1";;
2)
        PROV_POOL="choice 2";;
3)
        PROV_POOL="choice 3";;
*)
        STATUS=1
esac
if [ $STATUS == 0 ]
then
        echo "You entered a good value"
else
        echo "You have exceeded the error retry count."
fi
2

If you define your counter, it should get you past your initial problem:

COUNTER=0
3

I'm not great with tests so I don't know if this is the best way, but I had the same problem recently and finally got this to work best:

while  [[ $choice != '1' ]] && [[ $choice != '2' ]] && [[ $choice != '3' ]]; do
4

Just above the case put these statement

if [ -z "${choice}" ];then
        echo "Please specify the option\n"
        #Call the function here to loop it again.
fi
5

The quotes around the variable made the difference!