如果输入的输入被视为“无效”，则尝试继续请求输入

Question

这是我为 CS 课程制作的文字冒险游戏的一小部分。您正在探索一所房子，并通过告诉游戏您要向北、向南、向东还是向西来导航它。
因此，我想添加一些内容来告诉您，当您输入了无效输入时，如果您拼错了 Nroth、Suoth、Eas 或 Weast 之类的单词之一。这些只是示例，但希望您知道我的意思，只要它与北、南、东或西不完全匹配。 我将如何在这部分代码中做到这一点？

我做了一个错误示例，如果您犯了“elif room ==”porch”的拼写错误，我想输出的错误，但它应该继续询问您想去哪个方向，即使您收到该错误因为到目前为止，它会继续询问您想去哪个方向，并且无论您输入什么，它都不会根据您进入的房间输出应该说的文本。

def pickRoom(direction, room):
    if(direction == "quit") or (direction == "exit"):
        print("Better luck next time!")
        return "Exit"
    elif room == "Porch":
        if direction == "North":
            return "Pantry"
        else:
            print("That is not a valid entry!")
    elif room == "Pantry":
        if direction == "North":
            return "Kitchen"
        elif direction == "East":
            return "DiningRoom"
    elif room == "DiningRoom":
        if direction == "West":
            return "Pantry"
    elif room == "Kitchen":
        if direction == "West":
            return "LivingRoom"
        elif direction == "East":
            return "Bedroom"
    elif room ==  "Bedroom":
        if direction == "West":
            return "Kitchen"
    elif room == "LivingRoom":
        if direction == "West":
            return "Bathroom"
        elif direction == "North":
            return "Stairs"
    elif room == "Bathroom":
        if direction == "East":
            return "LivingRoom"
    elif room == "Stairs":
        if direction == "South":
            return "Bar"
    elif room == "Bar":
        if direction == "East":
            return "Shop"
    elif room == "Shop":
        if direction == "North":
            return "Closet"
        elif direction == "South":
            return "Storage"
    elif room == "Storage":
        if direction == "North":
            return "Shop"
    elif room == "Closet":
        if direction == "South":
            return "Shop"

如果您需要更大的代码部分甚至整个 .py 文件，请告诉我，谢谢。

Answer 1

为了保持在显示的代码部分内，根据要求，只需在末尾添加

else:
    print("Sorry, that does not make sense to me.")
    return room

这样可以解决当前问题，即如果没有与输入匹配的编程选项，则不可预测的值将被重新调整为新的当前房间。在这种情况下，通过返回参数room，存储当前房间的变量将继续使用有效房间（当前房间）。

当返回一个不可预测的值时，它很可能不是正确的房间名称之一，它需要这样才能使逻辑结构保持正常。一旦房间变量包含垃圾，它就不能再匹配任何选项，因此不会再输出任何有意义的东西。

作为防止房间变成垃圾的额外预防措施（由于任何可能的事故之一），您可以检查房间是否是现有房间之一，否则将其重置为默认值 - 或退出并显示错误消息 “糟糕，意外传送到未知空间。下次地球见。”

Answer 2

您可以尝试针对正确选择列表进行测试。我希望这会有所帮助！

if room in roomList:
    # availableDirection is a dictionary that 
    # has rooms as keys and valid directions as values.
    if direction in availableDirection[room]:
        # return the next room from a dictionary representing the 
        # current room where the key is direction and value is the next room.
    else:
        return "Invalid direction"
else:
    return "Invalid room"

Answer 3

我不确定您需要什么，但这可能会有所帮助：

directions = ["south", "west", "east", "north"]
while True:
    move = input("Choose which way you would like to go\n")
    if move.lower() in directions:
        print("You have chosen to go " + move)
    else:
        print("Invalid move!")

只要有一个想法，这是一个输出：

>>Choose which way you would like to go
north
>>You have chosen to go north
>>Choose which way you would like to go
North
>>You have chosen to go North
>>Choose which way you would like to go
nothr
>>Invalid move!
>>Choose which way you would like to go

Answer 4

elif room.lower() in ('perch','peach','pooch'):

你可以只列出一份你想指出的所有拼写错误的大清单。如果他们所做的选择不正确，请检查他们输入的值是否在此列表中。