Question

我尝试做一个简单的选择来选择一些东西。但我有错误，有人可以帮助我吗？

$ vi sel2.sh

lagi='y'

while  [ $lagi == 'y' ] || [ $lagi == 'Y' ];

do

clear

select menu in "Bakso" "Gado-Gado" "Exit";

   case $REPLY in 

        1) echo -n "Banyak mangkuk =";
           read jum
           let bayar=jum*1500;
           ;;
        2) echo -n "Banyak porsi =";
           read jum
           let bayar=jum*2000;
           ;;
        3) exit 0
           ;;
        *) echo "Sorry, tidak tersedia"
           ;;
  esac 
do

echo "Harga bayar = Rp. $bayar"
echo "THX"
echo 
echo -n "Hitung lagi (y/t) :";
read lagi;

    #untuk validasi input

    while  [ $lagi != 'y' ] && [ $lagi != 'Y' ] && [ $lagi != 't' ] && [ $lagi != 'T' ];
    do
       echo "Ops, isi lagi dengan (y/Y/t/Y)";
       echo -n "Hitung lagi (y/t) :";
       read lagi;
    done
done

Answer 1

尝试更改

select menu in "Bakso" "Gado-Gado" "Exit";

到

options=( "Bakso" "Gado-Gado" "Exit")
select menu in "${options[@]}"

Answer 2

您的代码中存在许多结构性问题。

可能这就是你要找的东西：

#!/bin/bash
lagi='y'

while  [ $lagi == 'y' ] || [ $lagi == 'Y' ]

do

    clear

    select menu in "Bakso" "Gado-Gado" "Exit"
    do
        case $REPLY in

            1) echo -n "Banyak mangkuk =";
               read jum
               let bayar=jum*1500;
               ;;
            2) echo -n "Banyak porsi =";
               read jum
               let bayar=jum*2000;
               ;;
            3) exit 0
               ;;
            *) echo "Sorry, tidak tersedia"
               ;;
        esac

        echo "Harga bayar = Rp. $bayar"
        echo "THX"
        echo
        echo -n "Hitung lagi (y/t) :";
        read lagi;

        #untuk validasi input

        while  [ $lagi != 'y' ] && [ $lagi != 'Y' ] && [ $lagi != 't' ] && [ $lagi != 'T' ];
        do
            echo "Ops, isi lagi dengan (y/Y/t/Y)";
            echo -n "Hitung lagi (y/t) :";
            read lagi;
        done
    done
done

错误选择Bash Linux上的菜单

2 个答案: