创建菜单和子菜单显示

Question

我有以下主菜单代码：

>>> def mainmenu ():
    d = ''
    while d == '':
        print ('\nM A I N   M E N U')
        print ('1. Settings')
        print ('q. Quit')
        option = input ('Select an option: ')
        if option.lower () == 'q':
            sys.exit ()
        elif option == '1':
            d = submenu ()
        else:
            print ('Invalid selection!')
    return d

>>>

子菜单代码：

>>> def submenu ():
    p = ''
    while p == '':
        print ('\nS U B  M E N U')
        print ('1. Perform an action')
        print ('b. Back')
        print ('q. Quit')
        option = input ('Select an option: ')
        if option.lower () == 'q':
            sys.exit ()
        elif option.lower () == 'b':
            mainmenu ()
            return
        elif option == '1':
            p = 'Do something'
        else:
            print ('Invalid selection!')
    return p

>>>

每当我从子菜单发出 "back" 命令返回主菜单时，整个菜单系统就会崩溃。看起来父菜单在它停止的地方继续调用另一个子菜单实例 where

p = ''

导致 value 为空。

下面是上述错误的示例：

>>> value = mainmenu ()

M A I N   M E N U
1. Settings
q. Quit
Select an option: 1

S U B  M E N U
1. Perform an action
b. Back
q. Quit
Select an option: b

M A I N   M E N U
1. Settings
q. Quit
Select an option: 1

S U B  M E N U
1. Perform an action
b. Back
q. Quit
Select an option: 1
>>> 
>>> value
>>>

但是，如果我从子菜单 select "option 1" 并且不使用 "back" 选项，一切都会按预期进行。

>>> value = mainmenu ()

M A I N   M E N U
1. Settings
q. Quit
Select an option: 1

S U B  M E N U
1. Perform an action
b. Back
q. Quit
Select an option: 1
>>> 
>>> value
'Do something'
>>>

如何修复 "back" 选项功能？

我试图在主菜单的选项 1 和我能想到的每个地方插入一个 return 语句，但它没有用。我缺乏一些逻辑，非常感谢任何解释或帮助。

...
    elif option == '1':
            d = submenu ()
            return
...

更新 1
- 添加变量 "msg" 到选项 == '1' 下的主菜单和 return 元组。

def mainmenu ():
    d = ''
    msg = ''    # Added in Update #1
    while d == '':
        print ('\nM A I N   M E N U')
        print ('1. Settings')
        print ('q. Quit')
        option = input ('Select an option: ')
        if option.lower () == 'q':
            sys.exit ()
        elif option == '1':
            msg = 'Option 1'    # Added in Update #1
            d = submenu ()
        else:
            print ('Invalid selection!')
    return msg, d    # Modified in Update #1

- 解包并打印 main ()

中的元组

message, action = mainmenu ()

print ('\nMessage: ', message)
print ('Action: ', action)

The answer below works but introduced another problem when the option 'b' is selected.

M A I N   M E N U
1. Settings
q. Quit
Select an option: 1

S U B  M E N U
1. Perform an action
b. Back
q. Quit
Select an option: b

M A I N   M E N U
1. Settings
q. Quit
Select an option: 1

S U B  M E N U
1. Perform an action
b. Back
q. Quit
Select an option: 1

Message:  Option 1
Action:  ('Option 1', 'Do something')   # <<<<<< Return a Tuple?

声明 return 一个元组，我已经从 main () 中解压了它。这是选项 'b' 未 selected.

时的工作示例

M A I N   M E N U
1. Settings
q. Quit
Select an option: 1

S U B  M E N U
1. Perform an action
b. Back
q. Quit
Select an option: 1

Message:  Option 1
Action:  Do something

Answer 1

发生这种情况是因为您 returning None。

elif option.lower () == 'b':
    mainmenu ()
    return

您的代码正在递归调用 mainmenu()，因此当您按 1 时，所有 return 值都会爬回到 "top," 并且在该过程中您引入了 NoneType.

只需return调用该函数即可。

elif option.lower () == 'b':
    return mainmenu ()

Answer 2

如前所述，问题是在子菜单中递归调用主菜单。换句话说，每次按下 "back" 键时，子菜单都会创建一个主菜单的新实例。要解决它，只需 return 到主菜单而不调用另一个主菜单。

来自子菜单：

elif option.lower () == 'b':
    return

此方法将 return "None" 到主菜单 while 循环将需要考虑到这一点。

def mainmenu ():
    d = None
    msg = ''
    while d is None:
    ....

创建菜单和子菜单显示

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