Bash: How to check in array check if input was entered twice? - bash

Question

The array variable ARRAY_ITEMS adding input from the user.
Here my script.
#!/bin/bash
var1=$(echo -e "Adding $INPUT to array.")
while true; do
printf "\n
A B C\n"
echo -e "This is the array: $ARRAY_ITEMS: "
read -p "Input: " INPUT
case "$INPUT" in
A) echo $var1 ; ARRAY_ITEMS+=$INPUT ;;
B) echo $var1 ; ARRAY_ITEMS+=$INPUT ;;
C) echo $var1 ; ARRAY_ITEMS+=$INPUT ;;
*) echo -e "Whoopsy! Invalid input." ;;
esac
done
The Output looks like this:
A B C This is the array: ABCAAABCAAA:
Input: D Whoopsy! Invalid input.
How does my script have to look like so that the following output appears for an input that has already been entered?
A B C
This is the array: ABC:
Input: A
ERROR. A is already selected.

Answer 1

Use pattern-matching to see if the input is already in the string.
items=
while true; do
printf "\n
A B C\n"
echo "This is the array: $items"
read -p "Input: " input
case "$INPUT" in
A|B|C) if [[ $items = *$input* ]]; then
echo "Error, $input is already in $items"
else
items+=$input
fi ;;
*) echo "Whoopsy! Invalid input." ;;
esac
done
If the input could be something more than a single letter, you'll need to use an array and resign yourself to walking the array, one element at a time, to compare them to the input. Something like
items=()
while true; do
...
found=0
case $input in
A|B|C) for x in "${items[#]}"; do
if [[ $x == $input ]]; then
echo "Error, already used $input"
found=1
break
fi
done
;;
*) echo "Error, invalid input" ;;
esac
(( found )) || items+=("$input")
done