如何修复我的 Python 函数，使其返回输入提示？

Question

我有一个菜单功能和选择功能都有效。有 3 种菜单选择。1 和 3 在某一时刻正常工作。2从来没有。我不知道我做了什么把它搞砸了，但是当我运行模块以通过 IDLE 进行测试时，它在第一次提示输入我的菜单选项编号之后就无法正常工作。它应该完成一个 if 语句，然后重新启动。我不知道还能尝试什么。我希望我知道我改变了什么来搞砸它。tribbles = 1modulus = 2closer= 3def menu():    print('    -MENU-')    print('1: Tribbles Exchange')    print('2: Odd or Even?')    print("3: I'm not in the mood...")menu()def choice():    choice = int(input('\n Enter the number of your menu choice: ')if choice == tribbles:    bars = int(input('\n How many bars of gold-pressed latinum do you have? '))    print('\n You can buy ',bars * 5000 / 1000,' Tribbles.')    menu()    choice()elif choice == modulus:    num = int(input('\n Enter any number:'))    o_e = num % 2    if num == 0:        print(num,' is an even number')    elif num == 1:        print(num,' is an odd number')    menu()    choice()elif choice == closer:    print('\n Thanks for playing!')    exit()else:    print('Invalid entry. Please try again...')    menu()    choice()print(' ')choice = int(input('\n Enter the number of your menu choice: '))我希望它返回字符串加上所有公式结果，然后再次询问，除非选择了选项 3 并执行了 exit()。但是，它在第一次输入后返回“输入您的菜单选项的编号：”，然后在第二个提示符上选择任何其他选项后返回空白。f

Answer 1

在检查 的值之前，未声明choice该变量。choice您必须在以下行之前捕获您的输入：if choice == tribbles:. 您只是在定义一个函数，它甚至不返回您选择的值或设置全局变量。

试试这个：

def menu():

    print('    -MENU-')

    print('1: Tribbles Exchange')

    print('2: Odd or Even?')

    print("3: I'm not in the mood...")

menu()

choice = int(input('\n Enter the number of your menu choice: '))

if choice == tribbles:

...

Answer 2

最好在文件顶部定义所有函数，并在底部调用这些函数！其次，您的缩进不正确，我假设您将其粘贴到此处后发生了这种情况。最后，您永远不会真正调用该函数choice()，而是用提示的结果覆盖它。

下面我将纠正这些问题。

tribbles = 1

modulus = 2

closer= 3

def menu():

    print('    -MENU-')

    print('1: Tribbles Exchange')

    print('2: Odd or Even?')

    print("3: I'm not in the mood...")

    choice() #added call to choice here because you always call choice after menu

def choice():

    my_choice = int(raw_input('\nEnter the number of your menu choice: ')) #you were missing a ) here! and you were overwriting the function choice again

    #changed choice var to my_choice everywhere

    if my_choice == tribbles:

        bars = int(raw_input('\nHow many bars of gold-pressed latinum do you have? '))

        print('\n You can buy ',bars * 5000 / 1000,' Tribbles.')

        menu()

    elif my_choice == modulus:

        num = int(raw_input('\n Enter any number:'))

        o_e = num % 2

        if num == 0:

            print(num,' is an even number')

        elif num == 1:

            print(num,' is an odd number')

        menu()

    elif choice == closer:

        print('\n Thanks for playing!')

        exit()

    else:

        print('Invalid entry. Please try again...')

        menu()

    print(' ')

if __name__ == "__main__": #standard way to begin. This makes sure this is being called from this file and not when being imported. And it looks pretty!

    menu()