while loop does not exit when the condition is not met

Question

When running the code below, I’d like to read a user-input answer until it is one of the 6 possible answers ([abcABC]), else continue with the next loop.

However, it does not exit the while loop when I enter one of the accepted answers.

I tried [] and [[]] for the conditions, I tried to place all of the conditions into one pair of square brackets, I tried to use | and ||, none of them worked as expected.

while [ "$ans" != "a" ] || [ "$ans" != "A" ] || [ "$ans" != "b" ] || \
      [ "$ans" != "B" ] || [ "$ans" != "c" ] || [ "$ans" != "C" ]; do
    read ans

    case $ans in
      [aA]) echo "aA"         ;;
      [bB]) echo "bB"         ;;
      [cC]) echo "cC"         ;;
      *)    echo "Try again." ;;
    esac
done

It should read in loop until one of the accepted answer is given; then it should continue with the following code (if any).

that other guy · Accepted Answer · 2019-05-21 17:35:05Z

2

Your loop condition is always true, and Bash correctly loops forever.

Consider a simplified version:

[ "$ans" != "a" ] || [ "$ans" != "b" ]

If ans is a then it becomes false || true which is true.
If ans is b then it becomes true || false which is true.
If ans is x then it becomes true || true which is true.

You wanted && instead of ||. Alternatively, do it in a single comparison: [[ "$ans" != [aAbBcC] ]]

answered May 21, 2019 at 17:35

that other guy

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1 Comment

tukusejssirs Over a year ago

This is the perfect answer! Yes, I see the reason. I have no idea why missed this. Anyway, thank you for the simplifacation --- I did not know I can use regex syntax instead of a string in comparison.

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