Question

当我在python中运行我的石头剪刀游戏时，故意将'rock'，'paper'或'scissors'中的一个拼错，正如预期的那样运行Player_num功能再次。但是，当我输入正确拼写的选项时，它返回数字，并将其作为NoneType返回;如果我第一次拼写正确，它会将变量编号作为int而不是NoneType返回。

我无法弄清楚如何解决这个问题，我试过跟踪变量，但我没有运气。

#Part of rock paper scissors game
def Player_num():
    #Player chooses one of rock paper or scissors
    print("Choose 'rock', 'paper', or 'scissors' by typing that word. ")
    guess = input()
    #The if statement is to decide whether the user's input is right or not
    if Valid_guess(guess):
        #if it is right, it continues with the game
        #the user's choice will be converted to a number 1,2 or 3
        if guess == 'rock':
            number = 1
        elif guess == 'paper':
            number = 2
        elif guess == 'scissors':
            number = 3
        return number
        #if the input is invalid, the system prompts the user to try it again
    else:
        print('That response is invalid.')
        Player_num()

#Part of rock paper scissors game
def Valid_guess(guess):
    #Validates the user's input
    if guess == 'rock' or guess == 'paper' or guess == 'scissors':
        status = True
    else:
        status = False
    #Returns the boolean value status
    return status

Answer 1

在您的函数结束时，在else块中，您已写入：

Player_num()

我认为你的意思是：

return Player_num()

否则，您将获得正确的输入，但不要将其返回给调用者。该函数反而运行，并返回None，默认返回值。

Answer 2

尝试使用以下内容：

def Player_num():
    print("Choose 'rock', 'paper', or 'scissors' by typing that word. ")
    guess = input()
    if Valid_guess(guess):
        if guess == 'rock':
            number = 1
        elif guess == 'paper':
            number = 2
        elif guess == 'scissors':
            number = 3
        return number
    else:
        print('That response is invalid.')
        return Player_num()

def Valid_guess(guess):
    if guess in ['rock', 'paper', 'scissors']:
        return True
    return False

Valid_guess也被简化为一个陈述。

返回仅在功能中第一次工作

2 个答案: