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python - 如何修复我的 Python 函数,使其返回时带有输入提示?

转载 作者:行者123 更新时间:2023-12-01 00:35:31 25 4
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我有一个菜单功能和选择功能,两者都有效。有 3 种菜单选择。 1 和 3 曾一度工作正常。 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不再工作了。它应该完成 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 = 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...")

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,除非选择选项 3 并执行 exit()。但是,在第一次输入后,它会返回“输入您的菜单选项的编号:”,然后在第二个提示上选择任何其他选项后,它会返回空白。f

最佳答案

要事第一!

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,我假设这是在您将其粘贴到此处之后发生的。最后,您实际上从未调用函数 choice(),而是用提示结果覆盖它。

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()

关于python - 如何修复我的 Python 函数,使其返回时带有输入提示?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/57814588/

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