在 bash 中创建子菜单
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Creating sub menu in bash
提问by Zenix
I'm trying to do the following :
我正在尝试执行以下操作：
function main_menu
{
option=0
until [ "$option" = "4" ]; do
echo " 1.) Add user"
echo " 2.) Remove user"
echo " 3.) Update user"
echo " 4.) Quit"
echo -n "Enter choice: "
read option
echo ""
case $option in
1 ) add_user ; press_enter ;;
2 ) remove_user ; press_enter ;;
3 ) update_user ; press_enter ;;
4 ) exit;;
* ) tput setf 4;echo "Please enter 1, 2, 3, or 4";tput setf 4;
esac
done
}
In a switch case(do while option is not 4).The 3rd option will show another sub menu with switch case as follows:
在 switch case 中（do while option is not 4）。第三个选项将显示另一个带有 switch case 的子菜单，如下所示：
function update_user
{
option=0
until [ "$option" = "3"]; do
echo " 1.) Update username"
echo " 2.) Update password"
echo " 3.) Return to menu"
echo -n "Enter choice: "
read option
echo ""
case $option in
1 ) update_username; press_enter ;;
2 ) update_password; press_enter ;;
3 ) main_menu; press_enter ;;
4 ) exit;;
* ) tput setf 3;echo "Please enter 1, 2 or 3";tput setf 3;
esac
done
}
The 3rd option goes back to the main menu, but when I try to quit the sub menu keeps appearing.
第三个选项返回到主菜单，但是当我尝试退出时，子菜单不断出现。
Anyone can advise me on a better way?
任何人都可以建议我更好的方法吗？
采纳答案by Cyrus
Replace two times $selectionwith $option, breakwith exit, remove }in function update_user, in function update_useradd }after doneand replace "$option" = "7"with "$option" = "4".
替换两次$selectionwith $option，breakwith exit，}在函数update_user中删除，在函数中update_user添加}之后done并替换"$option" = "7"为"$option" = "4"。
#!/bin/bash
function update_user
{
option=0
until [ "$option" = "3"]; do
echo " 1.) Update username"
echo " 2.) Update password"
echo " 3.) Return to menu"
echo -n "Enter choice: "
read option
echo ""
case $option in
1 ) update_username; press_enter ;;
2 ) update_password; press_enter ;;
3 ) main_menu; press_enter ;;
4 ) break ;;
* ) tput setf 3;echo "Please enter 1, 2 or 3";tput setf 3;
esac
# }
done
}
function main_menu
{
option=0
until [ "$option" = "4" ]; do
echo " 1.) Add user"
echo " 2.) Remove user"
echo " 3.) Update user"
echo " 4.) Quit"
echo -n "Enter choice: "
read option
echo ""
case $option in
1 ) add_user ; press_enter ;;
2 ) remove_user ; press_enter ;;
3 ) update_user ; press_enter ;;
4 ) exit;;
* ) tput setf 4;echo "Please enter 1, 2, 3, or 4";tput setf 4;
esac
done
}
main_menu
回答by John B
As pointed out by @KonstantinV.Salikhov, menus in bash are what selectloops are for.
正如@KonstantinV.Salikhov 所指出的，bash 中的菜单就是select循环的用途。
Here's one way you can implement your menu using select:
这是您可以使用以下方法实现菜单的一种方法select：
main_menu () {
options=(
"Add user"
"Remove user"
"Update user"
"Quit"
)
select option in "${options[@]}"; do
case $option in
${options[0]})
add_user
break
;;
${options[1]})
remove_user
break
;;
${options[2]})
update_user
break
;;
${options[3]})
exit
;;
*)
echo invalid option
;;
esac
done
}
main_menu
