python - python中函数之间的通信

Question

我正在做一个学校项目，我需要一些功能间沟通方面的帮助。这是我到目前为止得到的

def difficuilty():
level = 0
while level >=4 or level == 0:
    level = int(input("Please enter the difficulty (1/2/3)"))
    if level == 1:
        yesNo = input("you have chosen difficulty 1, is this correct? ")
        if yesNo.upper() == 'Y':
            level  = 1
        elif yesNo.upper() == 'N':
            level  = 4
        else:
            print ("You have entered the wrong thing")
    elif level == 2:
        yesNo = input("you have chosen difficulity 2, is this correct? ")
        if yesNo.upper() == 'Y':
            level  = 2
        elif yesNo.upper() == 'N':
            level  = 4
        else:
            print ("You have entered the wrong thing")
    elif level == 3:
        yesNo = input("you have chosen difficulity 3, is this correct? ")
        if yesNo.upper() == 'Y':
            level  = 3
        elif yesNo.upper() == 'N':
            level  = 4
        else:
            print ("You have entered the wrong thing")

return level 


def question(level):
    if level == 1:
        print ("hi")

def main():

    getName()
    difficulty()
    question(level)

我试图从难度函数中获取变量“level”以进入问题函数，以便我可以使用它，当我运行程序时，它给了我一个错误，上面写着“NameError: Name”level is not定义'。有人可以帮帮我吗。谢谢

Answer 1

变量level 只定义在你的两个函数的范围内，而不是在main() 的范围内。您必须在 main() 范围内定义一个变量(称为 level)才能访问它。尝试:

def main():
    getName()
    level = difficulty()
    question(level)

这样，从 difficulty() 返回的变量(在 difficulty() 中命名为 level)对于 是可访问的主()。

此外，这是您的实际代码吗？我注意到一些错误，比如 difficulty() 被两种不同的拼写方式，以及 difficuilty() 中缺少缩进，如果您运行它，将会产生意想不到的结果。如果您有逐字代码，请发布它，以便更容易判断具体问题是什么。