Multiple conditions in case statement, bash - linux

Question

read X;
read Y;
read Z;
if [[ $X,$Y,$Z -ne "0" ]]; then
if [[ $X,$Y,$Z -ge 1 && $X,$Y,$Z -le 1000 ]] && [[ $((X+Y)) -gt $Z || $((Y+Z)) -gt $X || $((X+Z)) -gt $Y ]]; then
case $X,$Y,$Z in
$X -eq $Y && $Y -eq $Z && $X -eq $Z ) echo "EQUILATERAL";;
esac
else
echo "bye";
fi
fi
*./bashtesting.sh: line 7: syntax error near unexpected token `-eq
./bashtesting.sh: line 7: $X -eq $Y && $Y -eq $Z && $X -eq $Z ) echo "EQUILATERAL";;*
how to equate the three variables at a time?

Answer 1

This is how I would do it, with a loop and associative array.
#!/usr/bin/env bash
for h in x y z; do
read -rp 'Enter input: ' input
if [[ -z $input ]]; then
printf >&2 '%s is empty, please try again!\n' "${input:-input}"
exit 1
elif [[ $input != *[0-9]* ]]; then
printf '%s is not an int, please try again!\n' "$input"
exit 1
else
declare -A num[$h]=$input
fi
done
for i in "${num[#]}"; do
if ! (( i == 0 )); then
for j in "${num[#]}"; do
if (( j > 1 && j < 100 )); then
if (( (num[x] + num[y]) > num[z] || (num[y] + num[z]) > num[x] || (num[x] + num[z]) > num[y] )); then
if (( num[x] == num[y] && num[y] == num[x] && num[x] == num[z] )); then
echo equilateral
break 2
fi
fi
fi
done
fi
done
It it not pretty, It looks like an Anti Pattern, and there might be a better way to do it but this will get you there I suppose.