Bash Linux上的错误选择菜单
我试着做一个简单的选择来选择一些东西。 但我有错误,有人能帮我吗 $vi sel2.shBash Linux上的错误选择菜单,linux,bash,select,case,Linux,Bash,Select,Case,我试着做一个简单的选择来选择一些东西。 但我有错误,有人能帮我吗 $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
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
试着改变
select menu in "Bakso" "Gado-Gado" "Exit";
到
代码中存在许多结构性问题 也许这就是你想要的:
#!/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
第7行错误请在此处张贴您的脚本,而不是it@mrid谢谢,很抱歉。请查看并帮助选择。您会遇到什么错误?
#!/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