在python

Question

我正在做一个学校项目，我需要一些帮助在功能之间进行通信。这是我到目前为止所得到的

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)

我正在尝试获取变量＆＃39; level＆＃39;从难度函数进入问题函数，所以我可以使用它，当我运行程序时，它给我一个错误，上面写着“NameError：Name＆＃39; level is not defined＆＃39 ;.有人可以帮帮我吗。感谢

Answer 1

变量Detected: x add(s)仅在两个函数的范围内定义，但不在level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：

main()

这样，def main(): getName() level = difficulty() question(level) 可以访问difficulty()（在level内名为difficulty()）返回的变量。

另外，这是您的实际代码吗？我注意到一些错误，例如main()拼写有两种不同的方式，而difficulty()内缺少缩进，如果你运行它会产生意想不到的结果。如果你有这个代码，请发布你的逐字代码，以便更容易分辨具体问题。

Answer 2

您需要检索从困难函数返回的级别值

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

Answer 3

我相信global方法可以做到这一点，它将函数中使用的局部arg转换为全局可调用的全局arg。

def difficuilty():
    level = 0
    while level >=4 or level == 0:
        level = int(input("Please enter the difficulty (1/2/3)"))
        global level

那种方式（级别）可以调用。最后一点，根据你的应用，我会选择Milo的答案，因为使用global方法可能有其缺点。