Question

我需要使'printmessage'函数起作用，以便打印动作消息，以便可以重复使用它，而不必多次使用打印。任何有帮助的事情都会很棒。我知道这很麻烦，但是如果可能的话，代码可以保持尽可能相似。现在，当我使用我的新代码时，会出现一条追溯错误消息

错误消息是：

Hello the four games avaliable are:
0   Mario Cart
1   Minecraft
2   Angry Birds
3   Grabd Theft Auto
What number game do you want? 1
You have chosen Minecraft

What is the level you would like to play at? 1
Begginer 
Traceback (most recent call last):
  File "/Users/megan/Desktop/Digital/Python/mock 1.py", line 87, in <module>
    main()
  File "/Users/megan/Desktop/Digital/Python/mock 1.py", line 85, in main
    actualmessage()
  File "/Users/megan/Desktop/Digital/Python/mock 1.py", line 72, in actualmessage
    print (action_message_header)
NameError: name 'action_message_header' is not defined

代码是：

#Ask user what game they would like to play
def game () :
    global gametype,gamelist
    gamelist = ["Mario Cart","Minecraft","Angry Birds","Grabd Theft Auto"]
    gamecount = 0
    LENGTH = len(gamelist)
    print ("Hello the four games avaliable are:")

    while gamecount < LENGTH:
        print (gamecount," ",gamelist[gamecount])
        gamecount = gamecount + 1

    while True:
        try:
            gamenum = int(input("What number game do you want? "))
            if 0 <= gamenum <= LENGTH:
                gametype = gamenum
                return gamenum
            print ("Please enter a number between 0 and",gamecount - 1,)
            print ("")

        except ValueError:
                print ("Please enter a valid value " )
                print ("")

gamenum = game ()
print ("You have chosen",gamelist[gamenum],)
print ("")

#Ask game level
def number():
    while True:
        try:
            Level = int(input("What is the level you would like to play at? "))

            if Level >= 1 and Level <= 25:
                print ("Begginer ")
                return ("Beginner")

            elif Level >=26 and Level <=75:
                print ("Intermediate")
                return ("Intermediate")

            elif Level >=76 and Level <=100:
                print ("Advanced")
                return ("Advanced")
            else:
                print("Out Of range(1-100): Please enter a valid number")
                print ("")

        except ValueError:
            print("Please enter a valid number")
            print ("")


#Create a subroutine to print out the action message
def printmessage ():
    action_message_header = """
    ("#                                                      #")
    ("########################################################")
    ("#################### ACTION MESSAGE ####################")
    ("########################################################")
    ("#                                                      #")
    """

def actualmessage ():
    global stationtype, num
    actual_message_header = """
    ("Play "+gamelist[gamenum]+" at"+num)
    """

    print (action_message_header)
    print (action_message_header)
    print (action_message_header)
    print (actual_message_header)
    print (actual_message_header)
    print (actual_message_header)

#This is to let the program work
def main ():
    global num
    num = number()

    printmessage ()
    actualmessage()

main()

Answer 1

您正在打印一个不返回任何值的函数。您可以通过将printmessage添加到下面的示例中来更改您的return action_message_header函数以返回要打印的值，然后继续操作

def printmessage ():
    action_message_header = """
    ("#                                                      #")
    ("########################################################")
    ("#################### ACTION MESSAGE ####################")
    ("########################################################")
    ("#                                                      #")
    """
    return action_message_header

或用以下内容替换您的printmessage函数：

def printmessage ():
    action_message_header = """
    ("#                                                      #")
    ("########################################################")
    ("#################### ACTION MESSAGE ####################")
    ("########################################################")
    ("#                                                      #")
    """
    print(action_message_header)

并调用printmessage()而不是打印函数。

当代码中包含#时，应使用三引号将它们全部读取为一个变量。

如何使用功能多次打印

1 个答案: