case语句中的多个条件，bash

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.