Question

所以，我的选择4似乎有问题。我希望它返回主菜单，主菜单可以由main（）调用。＆＃34;是＆＃34;显然有效，在输入＆＃34;是＆＃34;或＆＃34;是＆＃34;以外的任何内容时，应将其返回主菜单。输入＆＃34; no＆＃34;或者＆＃34; hhh＆＃34;只是退出程序以及＆＃34;是＆＃34; /＆＃34;是＆＃34;。它也显示相同的＆＃34;感谢您的演奏＆＃34;消息好像是＆＃34;是＆＃34;。我是Python的初学者，所以请耐心等待。

while endProgram == "Yes" or "yes":

    #Selection menu, user input to navigate.
    selection = eval(input("Your selection: "))

    #Selection 1, rules.
    if selection == 1:
        rpsRules()
        returnMain = input("\nWhen you would like to return to the Main Menu, press Enter.")
        main()

    #Selection 2, begin a match against the PC, calls againstPC module with choice as an argument.
    elif selection == 2:
        againstPC(choice)

    #Selection 3, begin a match against another player locally, calls twoPlayer module.
    elif selection == 3:
        twoPlayer()

    #Selection 4, end program, with an "Are you sure?" catch.    
    elif selection == 4:
        endProgram = input("\nAre you sure you want to quit? (Yes/No) ")
        if endProgram == "Yes" or "yes":
            print("\nThanks for playing Rock, Paper, Scissors!\nSee you next time!")
            break
        elif endProgram == "No" or "no":
            main()
        else:
            main()
    elif selection == 5:
        creatorCredits()

Answer 1

这一行：

while endProgram == "Yes" or "yes"

当你用英语读出来的时候，Doe并不意味着它的意思;你需要修改它，以便检查endProgram的两个值，如下所示：

while endProgram == "Yes" or endProgram == "yes"

然后你可以通过说：

进一步简化

while endProgram.lower() == "yes"

您永远不应该在用户输入上使用eval。这不仅是危险的，也是错误的重要来源，它可能会在您的应用程序中造成不可预测的行为。

事实上，您应该将此逻辑放在您打印菜单的main()方法中：

def main():
    # print the menu here
    selection = input("Your selection: ")
    try:
       selection = int(selection)
    except ValueError:
       print('Please enter a valid number')
       get_user_input()
    return selection

然后，在主要时间调用它：

while endProgram.lower() == "yes":
    selection = main()

您接下来的问题是本节，它与您的while循环进行相同的检查;这是不必要的。

elif selection == 4:
    endProgram = input("\nAre you sure you want to quit? (Yes/No) ")
    if endProgram == "Yes" or "yes":
        print("\nThanks for playing Rock, Paper, Scissors!\nSee you next time!")
        break
    elif endProgram == "No" or "no":
        main()
    else:
        main()

要清除所有这些，请按以下方式构建程序：

打印主菜单。
虽然主菜单中的选项不能退出，但请运行循环。
在循环中的每个选择之后，再次打印主菜单。

由于主菜单功能返回有效的响应;您也可以在那里移动退出逻辑：

def main():
    # print your menu here
    selection = input('Please enter your choice: ')
    try:
       selection = int(selection)
    except ValueError:
       print('Sorry, {} is not a valid choice'.format(selection))
       main()
    if selection == 4:
       exit_check = input('Are you sure you want to exit? Type Yes: ')
       return exit_check.lower()
    if 0 < selection < 4:
       print('{} is not a valid menu item.')
       main()
    return selection

现在，你的程序的主要逻辑循环是这样的：

end_program = main()
while end_program != 'yes':
     if end_program == 1:
         # do stuff
         end_program = main()
     if end_program == 2:
         # do stuff
         end_program = main()
print('Thank you, for playing. Good bye!')

Answer 2

答案是将您的IF语句更改为：

if endProgram == "Yes" or endProgram == "yes":
...

Do-While，菜单选择退出工作不正确

2 个答案: