将用户 input() 从 main() func 传递到下一个函数中的字符串

Question

我很难将我在主函数中输入提示的用户名传递到下一个函数中的字符串中。

我已经构建了一个函数程序，它将预先确定的名称传递到下一个函数字符串消息中。但是现在我已经尝试加强它并使用输入法，我在下一个函数中将它传递到字符串消息中非常费时。

def main():
    someMessage = input("Enter your name:")
    return someMessage

def buildGreeting (someMessage):
    message = "Greetings " +input(someMessage)+ " you have been hacked! This message will self destruct in ten seconds!"
    return message
def printMessage(aMessage):
    print(aMessage)

if __name__ == '__main__':
    main()

我想让它说“你好，Leif，你被黑了！这条消息将在十秒内自毁！”

这是我目前的结果。它提示输入我的名字，然后什么也不做。这是我运行程序时读取的内容。

输入你的名字：Leif

进程以退出代码 0 结束

Answer 1

另一种方法，

def main():
    someMessage = input("Enter your name:")
    return someMessage
    
someMessage = main()

def buildGreeting(someMessage):
    message = "Greetings " +someMessage+ " you have been hacked! This message will self destruct in ten seconds!"
    return message
    
aMessage = buildGreeting(someMessage)


def printMessage(aMessage):
    print(aMessage)
   

if __name__ == '__main__':
    printMessage(aMessage)

Answer 2

试试这个：

def main():
    someMessage = input("Enter your name:")
    buildGreeting(someMessage)

def buildGreeting (someMessage):
    message = "Greetings " +someMessage +" you have been hacked! This message will self destruct in ten seconds!"
    printMessage(message)

def printMessage(aMessage):
    print(aMessage)

if __name__ == '__main__':
    main()

buildGreeting 函数中的 +input(someMessage) 也不是必需的，因为您已经将该输入传递给该函数。

Answer 3

您会希望这样做，因为 main 需要调用其他函数才能打印任何内容：

def buildGreeting():
    name = input("Enter your name:")
    message = "Greetings " + name + " you have been hacked! This message will self destruct in 10 seconds."
    return message

def printMessage(aMessage):
    print(aMessage)

def main():
    message = buildGreeting()
    printMessage(message)

if __name__ == '__main__':
    main()

当我运行它时：

[dkennetz fun]$ python destruct.py
Enter your name:Dennis
Greetings Dennis you have been hacked! This message will self destruct in 10 seconds.

PS 消息不会自毁。