如何将函数链接到特定函数

Question

我想将某个函数链接到一个特定函数，该函数只有在前面的函数在 python 中成功执行时才会运行，并且链接的函数不应该在其他地方访问。

示例：我希望仅在名称函数成功运行时才运行问候函数。

def name():
    name = str(input("please enter your name: "))
    while name not in (""," "):
        return name
    def greeting():
        message = "hello {}, welcome".format(name)
        print(message)
        return message
    def happy():
        mood = str(input("hello {},this is to check if you are happy, kindly answer 'yes' or 'no': ".format(name))
        if mood == "yes":
            reply = "that's good"
        else:
            reply = "oh dear, cheer up"
        print(reply)
    else:
        break

info = name()
print(info)

这段代码应该只在name函数执行成功的情况下运行greeting函数并调用happy函数，并且如果没有name函数也不应该调用greeting或happy函数。

但它没有按预期执行。它只返回名称。

提前致谢

Answer 1

在name() 函数中定义其他函数有点不必要。我建议单独定义函数，然后在下面的 while 循环中相应地调用它们。

def name():
    name = str(input("please enter your name: "))
    if name not in (""," "):
        return name
    return None

def greeting(name):
    message = "hello {}, welcome".format(name)
    print(message)

def happy(name):
    mood = input("hello {},this is to check if you are happy, kindly answer 'yes' or 'no': ".format(name))
    if mood == "yes":
        reply = "that's good"
    else:
        reply = "oh dear, cheer up"
    print(reply)

while True:
    info = name()
    if info:
        greeting(info)
        happy(info)
        break

编辑：我意识到您不希望在 name 函数范围之外访问其他函数。也可以这样做（虽然很奇怪）。

def name():
    name_ = str(input("please enter your name: "))
    if name_ not in (""," "):
        def greeting(name):
            message = "hello {}, welcome".format(name)
            print(message)

        def happy(name):
            mood = input("hello {},this is to check if you are happy, kindly answer 'yes' or 'no': ".format(name))
            if mood == "yes":
                reply = "that's good"
            else:
                reply = "oh dear, cheer up"
            print(reply)
            return mood

        greeting(name_)
        return name_, happy(name_)

    return name()

info = name()
print(info)