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我在让我的代码正确“返回”选项字符串之一时遇到问题。

如果用户第一次正确输入了其中一个字符串选项，那么在执行“while”到“if/elif”语句时，“return”值会完美返回。没问题。

但是，如果用户第一次没有正确输入数据，我会尝试用我最后的“else”语句来捕捉它，然后再次开始该函数。然而，第二次/第三次/等时间，即使用户输入了一个有效的选择，'return' 类型是 None 并且返回的值是 None。

所以，我的用户验证缺少一些东西。有什么想法吗？

#Global variable initialization
mainSelection = str(None)

#Santa's Christmas Card List Main Menu function
def XmasMainMenu(mainSelection):
    print('')
    print('How may the elves be of service today?')
    print('')
    print('\"PRINT\"  - Print out the Christmas card list from the database.')
    print('\"ADD\"    - Add nice recipients information to the Christmas card list.')
    print('\"SEARCH\" - Search the database for information.')
    print('\"DELETE\" - Remove naughty recipients from the Christmas card list. ')
    print('\"EXIT\"   - Exit the program without any printing or changes to the database.')
    print('')
    mainSelection = input('Please enter your selection: ')
    mainSelection = mainSelection.lower()
    print(type(mainSelection), mainSelection)

#Selection return value
    while mainSelection != None:
        if mainSelection == 'print':
            print('|| Will print out the Xmas Card List ||')
            return mainSelection
        elif mainSelection == 'add':
            print('|| Will prompt user to add information to the DB ||')
            return mainSelection
        elif mainSelection == 'search':
            print('|| Will prompt user to search the information in the DB ||')
            return mainSelection
        elif mainSelection == 'delete':
            print('|| Will prompt the user to delete recipients from the DB ||')
            return mainSelection
        elif mainSelection == 'exit':
            print('|| Will exit the user from the program with no changes')
            return mainSelection
        elif mainSelection == 'drop table':
            print('|| Will call the XmasTableDrop function ||')
            return mainSelection
        else:
            print('')
            print('Input Error:  Please enter a valid selection above!')
            print('Try again...')
            print('')
            print(type(mainSelection), mainSelection)
            break
    XmasMainMenu(mainSelection)

程序启动

用户输入正确，“返回”值正确

第一个用户输入无效。收到错误消息，功能重新开始。第二个用户输入有效->但是，“返回”类型为无，值为无（这是我需要修复的，但无法弄清楚）。

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1 回答 1

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这是典型的“新程序员”错误。除非您知道递归是什么以及它是如何工作的，否则不要在函数本身内调用函数。当被调用函数返回时，它会返回调用它的函数。会发生什么：

XmasMainMenu is called from the top-level script.
     ...user enters an incorrect value...
     XmasMainMenu calls XmasMainMenu again.
         ...user enters a correct value....
         XmasMainMenu (2nd call) returns to XmasMainMenu (1st call)
     The return value is not assigned to anything, so is lost.
     Now at the end of the function with no return, so returns default None.
top-level script receives None.

相反，将您的代码包装在 a 中while True:，break当您获得正确的值时，然后返回该值（伪代码）：

def XmasMainMenu():

    while True:

       print menu  # this could be outside while if you don't want to re-print menu.
       selection = input()

       if selection valid:
           break  # exit while loop
       else:
           prompt user to enter valid input.

    return selection
于 2020-11-30T23:36:11.003 回答