linux - Bash Linux上的错误选择菜单

我试图做出一个简单的选择来选择一些东西。
但是我有错误，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

最佳答案

尝试改变

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

至

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

关于linux - Bash Linux上的错误选择菜单，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/47118857/