檢查輸入是否為 integer 並等於 bash 中的特定數字

Question

嘗試檢查用戶輸入的值是否是 integer 以及等於 35 或 75。如果輸入的值是 integer 並且等於 35 或 75，那么我想通知用戶這一點。 如果輸入的值不是 35 或 75（例如字符串、null 或不同的整數），那么我的目標是讓用戶知道它無效並讓他們重試。

read -p 'Please enter an integer that is either equal to 35 or equal to 75: ' value

if [ $value =~ ^[0-9]+$ ] && ([ $value -eq 35 ] || [ $value -eq 75 ]
    echo "The input is acceptable"
    exit 1

else
    echo "The value is invalid. Try again."

fi

exit 0

我一直收到的錯誤是第 10 行：意外標記“else”附近的語法錯誤

Answer 1

你錯過了then 。 我會這樣整理：

#!/bin/bash

read -r -p 'Please enter an integer that is either equal to 35 or equal to 75: ' value

if [[ "$value" =~ ^[0-9]+$ ]] && [ "$value" -eq 35 ] || [ "$value" -eq 75 ]
    then
    echo "The input is acceptable"
    exit 0

else
    echo "The value is invalid. Try again."

fi

exit 0

更新：為了不斷提示用戶輸入，請執行以下操作：

#!/bin/bash

while 
  read -r -p 'Please enter an integer that is either equal to 35 or equal to 75: ' value
do
if [[ "$value" =~ ^[0-9]+$ ]] && [ "$value" -eq 35 ] || [ "$value" -eq 75 ]
    then
    echo "The input is acceptable"
    exit 0
else
    echo "The value is invalid. Try again."

fi
done

exit 0

Answer 2

當答案正確時不確定exit 1 ，我猜你想做這樣的事情：

#!/bin/bash

read -p 'Please enter an integer that is either equal to 35 or equal to 75: ' value

if [[ $value =~ ^[0-9]+$ && ($value -eq 35 || $value -eq 75) ]]; then
  echo "The input is acceptable"
else
  echo "The value is invalid. Try again."
  exit 1
fi

exit 0

我個人會將其簡化為：

#!/bin/bash

read -p 'Please enter an integer that is either equal to 35 or equal to 75: ' value

if [[ "${value}" == "35" || "${value}" == "75" ]]; then
  echo "The input is acceptable"
else
  echo "The value is invalid. Try again."
  exit 1
fi

exit 0