再次播放功能导致在刽子手游戏中被要求再次播放的循环

Question

我有一个简单的重玩功能，在游戏结束时（在用户输赢之后）他们可以选择重玩。然而，如果用户没有键入 'Y'（是）或 'N'（否），那么他们应该得到消息 'Please type a valid answer.'。这是行不通的。实际发生的情况是，用户在输入任何字符（甚至是或否）后，都会陷入被要求再次玩游戏的循环中。

        playagain = input('Wanna play again? (Y/N) ').upper()
        if playagain == 'Y':
            word = random.choice(wordbank)
        elif playagain == 'N':
            print('Alright, goodbye.')
        while playagain != 'Y' or 'N':
            print('Please type a valid answer. (Y/N)')
            print()
            playagain = input('Wanna play again? (Y/N) ').upper()

如果我删除“或 'N'”，则再次播放的选项有效，但前提是用户输入是。如果他们输入 no，那么他们就会遇到再次被要求玩的循环。我该如何解决这个问题？

这是我的全部代码：

# importing wordbank
import random
from wordbankcool import wordbank

# hangman graphics
hangman_graphics = ['_',
                    '__',
                    '__\n |',
                    '__\n |\n O',
                    '__\n |\n O\n |',
                    '__\n |\n O\n/|',
                    '__\n |\n O\n/|\ ',
                    '__\n |\n O\n/|\ \n/',
                    '__\n |\n O\n/|\ \n/ \ '
                    ]

# code is inside while loop
playagain = 'Y'
while playagain == 'Y':

    # alphabet
    alphabet = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l',
                'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z']

    # basic functions of the game
    mistakes = 0
    letters_guessed = []
    mistakes_allowed = len(hangman_graphics)

    # selecting a random word for the user to guess
    word = random.choice(wordbank)

    # letters user has guessed stored in list, making each letter of the word seperate
    letters_word = list(word)
    wrong_letters = []

    print()

    # amount of letters the word has
    print(f'The word has {len(letters_word)} letters')

    # while loop which will run until the the number of mistakes = number of mistakes allowed
    while mistakes < mistakes_allowed:
        print()
        print('Incorrect guesses: ', end='')
        for letter in wrong_letters:
            print(f'{letter}, ', end='')
        print()
        print(f'Guesses left: {mistakes_allowed - mistakes}')
        letter_user = input('Guess a letter: ').lower()

    # checking if the letter has been guessed before
        while letter_user in letters_guessed or letter_user in wrong_letters:
            print()
            print('You have already guessed this letter. Please guess a different one.')
            letter_user = input('Guess a letter: ')

    # increasing amount of mistakes if the letter that has been guessed is not in the word + checking if guess is a letter
        if letter_user not in alphabet:
            print('Please only enter A LETTER.')
            continue
        if letter_user not in letters_word:
            mistakes += 1
            wrong_letters.append(letter_user)

        print()

        # showing how many letters the user has/has not guessed
        print('Word: ', end='')

    # if letter is in word, its added to letters guessed
        for letter in letters_word:
            if letter_user == letter:
                letters_guessed.append(letter_user)

    # replace letters that haven't been guessed with an underscore
        for letter in letters_word:
            if letter in letters_guessed:
                print(letter + ' ', end='')
            else:
                print('_ ', end='')

        print()

    # hangman graphics correlate with amount of mistakes made
        if mistakes:
            print(hangman_graphics[mistakes - 1])
        print()
        print('-------------------------------------------')  # seperator

    # ending: user wins
        if len(letters_guessed) == len(letters_word):
            print()
            print(f'You won! The word was {word}!')
            print()
            playagain = input('Wanna play again? (Y/N) ').upper()
            if playagain == 'Y':
                word = random.choice(wordbank)
            elif playagain == 'N':
                print('Alright, goodbye.')
            while playagain != 'Y' or 'N':
                print('Please type a valid answer. (Y/N)')
                print()
                playagain = input('Wanna play again? (Y/N) ').upper()

    # ending: user loses
    if mistakes == mistakes_allowed:
        print()
        print('Unlucky, better luck next time.')
        print()
        print(f'The word was {word}.')
        print()
        playagain = input('Wanna play again? (Y/N) ').upper()
        if playagain == 'Y':
            word = random.choice(wordbank)
        elif playagain == 'N':
            print('Alright, goodbye.')
        while playagain != 'Y' or 'N':
            print('Please type a valid answer. (Y/N)')
            print()
            playagain = input('Wanna play again? (Y/N) ').upper()

Answer 1

您目前的情况：

while playagain != 'Y' or 'N':

or 运算符需要两个布尔值。所以 Python 将尝试将 'N' 转换为布尔值。由于 'N' 是一个非空字符串，它的计算结果将是 True。因此，您所拥有的条件将等同于：

while playagain != 'Y' or True:

这将始终计算为 True。

要解决此问题，while 条件应如下所示。请注意，我们还将 or 替换为 and，以使逻辑正常。

while playagain != 'Y' and playagain != 'N':

Answer 2

我们必须为两者写下完整的条件。尝试这样做：

while playagain!='Y' and playagain!='N':

再次播放功能导致在刽子手游戏中被要求再次播放的循环

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