[FIXED] How to create a menu with a range of options?

Issue

I would like to create a menu in shell script with a range of options. In this case, I want to force the user only to use a range of numbers, e.g. 1 to 5, but without use CASE. If the user choose 6, the menu ask again for the number between the range.

I remember something like:

OPTION (){
    [[ $option = +(1|2|3|4|5) ]] || OPTION
}

Solution

Following may help you in same:

cat choose2.ksh
check() {
while [ ! ${finished} ]
do
echo "Please enter a digit:"
read value
if [[ $value -le 5 ]]
then
        echo "user entered between 1 to 5."
        finished=1
else
        echo "user entered more than 5 in this case."
fi
done
}

check

Execution of script:

./choose2.ksh
Please enter a digit:
12
user entered more than 5 in this case.
Please enter a digit:
12
user entered more than 5 in this case.
Please enter a digit:
1
user entered between 1 to 5.

So you could see that if user enters other than 1 to 5 than it asks user again a Input else it will simple come out of script(you could do other things too as per your need).

Answered By - RavinderSingh13

Answer Checked By - Marilyn (PHPFixing Volunteer)