Python摇滚剪刀虫我不明白

Question

我正在用Python编写一个摇滚剪刀游戏，我对下面的结果非常满意，直到我注意到我得到了下面的“无效答案”的情况：

player1 =p和player2 =s

player1 =s和player2 =r

我不明白为什么对于下面的代码逻辑，这些情况不起作用。同样字母的其他组合对两位玩家都很好。

我做错什么了？

(我在用朱庇特做蟒蛇)

scoreP1 = 0
scoreP2 = 0
game = True
while game != 'QUIT':
    player1 = input("\nJanken! Player 1, type 'r' for Rock, 'p' for Paper or 's' for Scissor: ")
    player2 = input("Janken! Player 2, type 'r' for Rock, 'p' for Paper or 's' for Scissor: ")
    if(
        (player1=="r" and player2=="s") or
        (player1=="p" and player2=="r") or
        (player1=="s" and player2=="p")
    ):
        print("\nPlayer 1 wins! Congrats! Updating scores...")
        scoreP1+=1
        print("Score player 1:",scoreP1)
        print("Score player 2:",scoreP2)
    elif(
        player1==player2
    ):
        print("\nTie! Updating scores...")
        print("Score player 1:",scoreP1)
        print("Score player 2:",scoreP2)
        
    elif(
        player1 != "r" or
        player1 != "s" or
        player1 != "p" or
        player2 != "r" or
        player2 != "s" or
        player2 != "p"
    ):
        print("\nNot a valid answer. Please type 'r' for Rock, 'p' for Paper or 's' for Scissor.")
    else:
        print("Player 2 wins! Congrats! Updating scores...")
        scoreP2+=1
        print("Score player 1:",scoreP1)
        print("Score player 2:",scoreP2)
    game = input("\nAnother game? Type 'QUIT' to end this, or anything else to continue:")

Answer 1

问题是，对于应该返回"Player 2 wins“(例如player1=="s”和player2=="r")的结果，无效的if块匹配：

player1 != "r" or
player1 != "s" or
player1 != "p" or
player2 != "r" or
player2 != "s" or
player2 != "p"

你可能是说

( player1 != "r" and player1 != "s" and player1 != "p" ) 
or ( player2 != "r" and player2 != "s" and player2 != "p" )

或者，更多的丙酮：

 player1 not in ["s", "p", "r"] or player2 not in ["s", "p", "r"] 

Answer 2

elif(
    player1 != "r" or
    player1 != "s" or
    player1 != "p" or
    player2 != "r" or
    player2 != "s" or
    player2 != "p"
):

这个代码块需要一行为True，才能使整个elif变为true。

如果是player1 == 's'，代码会说"s不等于r“，因此‘or’-语句的列表必须是True。

Answer 3

在最后一个elif中，您正在测试player1或player2是否与您接受的答案不同。

if (player1 != "r" or player1 != "s"...)

这是真的，因为如果它是"r“，它不是"s"，如果是"s”，那么它就是!= "r“等等。

如果和不属于任何一个thiese，那么在测试时应该使用它而不是OR。用和和替换那些或，你应该会没事的。