制作石头、布、剪刀

Question

我在做一个石头，布，剪刀的游戏。你知道吗

下面的代码可以工作，但缺少两个关键的东西，我非常想知道如何实现这一点：

1. 如果是平局，选手们必须能够重新选择。所以break语句必须返回，但是需要什么才能返回while循环并再次播放呢？

2. 比赛结束时也是如此。球员必须能够再次比赛。我想它和上面的代码是一样的。

答案将非常有用，因为它真的帮助我了解while循环更多。你知道吗

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

这段代码应该能满足你的需要

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? ')

The code can be optimised and simplified by doing away few if and else, but I kept this for your understanding and making minor tweak in your code.