Question

我正在做一个石头剪刀布游戏。

下面的代码有效，但是缺少两个关键的东西，我真的很想知道如何实现这一点：

如果出现平局，玩家必须能够再次选择。因此break语句必须执行，但是返回while循环并再次播放需要什么？
游戏结束时也是如此。玩家必须能够再次玩。我认为这与上面的代码相同。

答案非常有帮助，因为它确实可以帮助我更多地了解while循环。

def rps():

    play = True

    player1 = input('Player1: rock, paper of scissors? ')
    player2 = input('Player2: rock, paper of scissors? ')

    while play:
        if player1 == 'rock' and player2 == 'rock': # deze methode werkt niet om opnieuw weer te beginnen
            print ('Tie! Pick again')
            break
        elif player1 == 'rock' and player2 == 'paper':
            print('Player 2 won')
            break
        elif player1 == 'rock' and player2 == 'scissors':
            print ('Speler 1 won')
            break
        elif player1 == 'paper' and player2 == 'rock':
            print('Speler 1 won')
            break
        elif player1 == 'paper' and player2 == 'paper':
            print('Tie! Pick again')
            continue
        elif player1 == 'paper' and player2 == 'scissors':
            print('Speler 2 won')
            break
        elif player1 == 'scissors' and player2 == 'rock':
            print('Speler 2 won')
            break
        elif player1 == 'scissors' and player2 == 'paper':
            print('Speler 1 won')
            break
        elif player1 == 'scissors' and player2 == 'scissors':
            print('Tie! Pick again')

Answer 1

在第二个while循环中包装游戏性循环。不要打断关系。游戏循环的输入发生在游戏循环中。退出或玩法的输入发生，并在游戏循环之前进行评估。

while True:
    print("Welcome to Rock, Paper, Scissors.")
    gameplay = input("(P)lay or (Q)uit?")
        if (gameplay == 'Q'):
             break
        else:
             while True:
                player1 = input('Player1: rock, paper of scissors? ')
                player2 = input('Player2: rock, paper of scissors? ')
                {Gameplay Logic}

Answer 2

此代码应能解决您的目的

def rps():

    play = True

    player1 = input('Player1: rock, paper of scissors? ')
    player2 = input('Player2: rock, paper of scissors? ')

    while play:
        if player1 == 'rock' and player2 == 'rock': # deze methode werkt niet om opnieuw weer te beginnen
            print ('Tie! Pick again')
        elif player1 == 'rock' and player2 == 'paper':
            print('Player 2 won')
            if input('Type c to continue playing, anything else to quit') == 'y':
                pass
            else:
                break
        elif player1 == 'rock' and player2 == 'scissors':
            print ('Speler 1 won')
            if input('Type c to continue playing, anything else to quit') == 'c':
                pass
            else:
                break
        elif player1 == 'paper' and player2 == 'rock':
            print('Speler 1 won')
            if input('Type c to continue playing, anything else to quit') == 'c':
                pass
            else:
                break
        elif player1 == 'paper' and player2 == 'paper':
            print('Tie! Pick again')

        elif player1 == 'paper' and player2 == 'scissors':
            print('Speler 2 won')
            if input('Type c to continue playing, anything else to quit') == 'c':
                pass
            else:
                break
        elif player1 == 'scissors' and player2 == 'rock':
            print('Speler 2 won')
            if input('Type c to continue playing, anything else to quit') == 'c':
                pass
            else:
                break
        elif player1 == 'scissors' and player2 == 'paper':
            print('Speler 1 won')
            if input('Type c to continue playing, anything else to quit') == 'c':
                pass
            else:
                break
        elif player1 == 'scissors' and player2 == 'scissors':
            print('Tie! Pick again')

        player1 = input('Player1: rock, paper of scissors? ')
        player2 = input('Player2: rock, paper of scissors? ')

只需减少一些if和否则，但我保留了此信息是为了您的理解并进行一些细微调整您的代码。

制作石头剪刀布游戏

2 个答案: