如果键不在字典中，则运行 while 循环

Question

我目前正在尝试构建一个非常简单的程序，在该程序中要求用户选择路径并根据选择的路径来更新字典。似乎当用户首先选择 path_2b 时，它继续前进并打印“恭喜完成游戏”并停止运行，即使只有 path_2b 已添加到字典中，但它应该只在两条路径都存在时才停止运行。我只是开始学习 Python 的基本知识和一般的编程知识，因此感谢您提供任何帮助和提示！

user_save = {}

def start_button():
    def path_2a():
        if "path_1" in user_save:
            print("You've already taken this path.")
        else:
            user_save["path_1"] = "completed"
            print("Congrats on finishing this path!")
    def path_2b():
        if "path_2" in user_save:
            print("You've alredy taken this path.")
        else:
            user_save["path_2"] = "Completed"
            print("Congrats on finishing this path!")

    chosen_path = input("Would you like to choose path 2A or 2B?: ").lower()
    if chosen_path == "2a":
        path_2a()
    elif chosen_path == "2b":
        path_2b()
    else:
        print("Sorry that isn't a valid path. Please try again.")

while ("path_1" and "path_2") not in user_save:
    start_button()
if "path_1" and "path_2" in user_save:
    print("Congrats on finishing the game!")

我希望循环继续运行，直到用户选择了路径 1 和 2。一旦两个键都在字典中，我想打印一条祝贺消息并打破循环。就像我之前说的大多数代码运行良好。如果用户先选择 path_2a 然后选择 2b，则即使他们选择不存在的路径，循环也会完全按照我的意愿进行。只有当用户首先选择 path_2b 时。感谢您的帮助！

Answer 1

根据您的代码，您需要更改代码：

user_save = {}

def start_button():
    def path_2a():
        if "path_1" in user_save:
            print("You've already taken this path.")
        else:
            user_save["path_1"] = "completed"
            print("Congrats on finishing this path!")
    def path_2b():
        if "path_2" in user_save:
            print("You've alredy taken this path.")
        else:
            user_save["path_2"] = "Completed"
            print("Congrats on finishing this path!")

    chosen_path = input("Would you like to choose path 2A or 2B?: ").lower()
    if chosen_path == "2a":
        path_2a()
    elif chosen_path == "2b":
        path_2b()
    else:
        print("Sorry that isn't a valid path. Please try again.")

while  "path_2" not in user_save or "path_1" not in user_save:
    start_button()

if "path_2" in user_save and "path_1" in user_save:
    print("Congrats on finishing the game!")

Answer 2

不要将函数放在函数中，而是尝试以下操作。我已将其简化为使用“2a”和“2b”作为实际的键名。此外，由于 2a 和 2b 都决定了您想要退出的时间，因此您可以跳过 while 循环并只使用一个函数，该函数在选择 2a 和 2b 时退出，否则继续运行。

def play_game(user_save):
    chosen_path = input("Would you like to choose path 2A or 2B?: ").lower()

    if chosen_path not in ['2a', '2b']:
        print("Sorry that isn't a valid path. Please try again.")
    elif chosen_path in user_save:
        print("You've already taken this path.")
    else:
        user_save[chosen_path] = "Completed"
        print("Congrats on finishing this path!")

    # If both 2a and 2b have been taken, just "return" to exit.
    if '2a' in user_save and '2b' in user_save:
        print("Congrats on finishing the game!")
        return

    # If they're still here, loop it again. Take the updated
    # dictionary <user_save> as input for the next round.
    play_game(user_save)


# Run the program
user_save = {}
play_game(user_save)