Bash 不會停止第二個“case”語句

Question

在下面的腳本中，如果我在整個腳本中給出y 。 在第二種case $yn in的情況下，它不會要求 y/n。 它只要求 y/n 一次而不是兩次。

#!/bin/bash

echo -e "\nFirst Step?"

while true; do
    case $yn in
        [Yy]* ) echo "going through first"; break;;
        [Nn]* ) exit;;
        * ) read -p "Please answer yes or no: " yn;;
    esac
done

echo -e "\nSecond Step?"

while true; do
    case $yn in
        [Yy]* ) echo "going through second"; break;;
        [Nn]* ) exit;;
        * ) read -p "Please answer yes or no: " yn;;
    esac
done

如何解決這個問題呢。

Answer 1

在每個循環中將read命令移到 case 語句的上方。

shopt -s nocasematch

printf '\nFirst step\n'

while true; do
    read -p "Please answer yes or no: " yn
    case $yn in
        y* ) echo "going through first"; break;;
        n* ) exit;;
    esac
done


printf '\nSecond step\n'

while true; do
    read -p "Please answer yes or no: " yn
    case $yn in
        y* ) echo "going through second"; break;;
        n* ) exit;;
    esac
done

或者，使用select ：它為您處理循環，並將變量設置為已知值

PS3="Please answer Yes or No: "
select ans in Yes No; do
    case $ans in
        Yes) echo "going through nth"; break ;;
        No) exit
    esac
done

Answer 2

冒着明顯的危險： yn已經從第一個案例步驟中設置，因此第二次不需要將 go 設置為默認案例並再次讀取輸入。

只需將第二個yn重命名為yn2或類似名稱。

#!/bin/bash

echo -e "\nFirst Step?"

while true; do
    case $yn in
        [Yy]* ) echo "going through first"; break;;
        [Nn]* ) exit;;
        * ) read -p "Please answer yes or no: " yn;;
    esac
done

echo -e "\nSecond Step?"

while true; do
    case $yn2 in
        [Yy]* ) echo "going through second"; break;;
        [Nn]* ) exit;;
        * ) read -p "Please answer yes or no: " yn2;;
    esac
done