如何從python中的兩個if語句接受答案

Question

我對python代碼中的嵌套if語句有疑問。 我想知道我是否可以在while循環中有兩個if語句，並且兩個語句都在內部嵌套一個if語句。 這樣，該人員將能夠從嵌套的if語句中進行回復，或者從while循環中的if語句中進行回復。 對我的代碼的任何想法將不勝感激。

這是我嘗試制作的游戲的代碼示例。 我正在嘗試學習並將新技能納入其中，因此這是我的問題。

有沒有一種方法可以確保當我在問題輸入中鍵入“向前看”時，對於第二個輸入，我說“向后看”而不是“去那里”，那么該輸入可能是問題輸入而不是move_on輸入？

def first_room():
    print("Your in a room how to get out...")
    next_room = True
    while next_room:
        question = input()
        if question == "look forward".lower():
            print("You are looking forward")
            move_on = input()
            if move_on == 'go there'.lower():
                next_room = False
                second_room()
            else:
                pass
        if question == 'look backward'.lower():
            print("You are looking backward")
            move_on = input()
            if move_on == 'go there'.lower():
                next_room = False
                third_room()
            else:
                pass


def second_room():
    print("This is the second room")


def third_room():
    print("This is the third room") 


first_room()

Answer 1

我可以測試，但我認為可能是這樣

我使用帶有True/False變量來記憶游戲的"state"

順便說一句：如果您在另一個功能中運行一個房間，您將具有隱藏的“遞歸”。 最好返回函數名稱（不帶() ）並在函數外部運行它。

def first_room():
    print("Your in a room how to get out...")

    looking_forward = False
    looking_backward = False

    while True:
        question = input().lower()
        if question == "look forward":
            if not looking_forward:
                print("You are looking forward")
                looking_forward = True
                looking_backward = False
            else:            
                print("You are already looking forward, so what next")
        elif question == 'look backward':
            if not looking_backward:
                print("You are looking backward")
                looking_forward = False
                looking_backward = True
            else:            
                print("You are already looking backward, so what next")
        elif question == 'go there':
            if looking_forward:
                return second_room # return function name to execute outside
            if looking_backward:
                return third_room # return function name to execute outside
            else:
                print('Go where ???')


def second_room():
    print("This is the second room")


def third_room():
    print("This is the third room") 

# --------------------------------

next_room = first_room # assign function without ()

while next_room is not None:
    next_room = next_room() # get function returned from function 

print("Good Bye")    

Answer 2

受到furas關於遞歸的評論的啟發，我想知道如何處理具有通用功能的任何房間。 在當前設計中，許多房間需要許多功能，這些功能將一遍又一遍地使用許多相同的代碼。 如果每個房間在某種程度上都是“特殊的”，需要特定的代碼，這可能是不可避免的。 另一方面，如果您有一種通用的方式來瀏覽房間，則可以節省很多工作。

所以我在想，也許您可​​以以某種常見的形式存儲房間屬性，並具有一個一般功能，要求玩家輸入並建議他們可以做什么。 作為“通用格式”，我在這里選擇了一個詞典列表，確定它是最容易使用的詞典。

每個列表索引（零索引為+1）對應一個房間號，例如roomlist[0]是第一個房間。 每個字典都存儲了我們需要了解的有關房間的信息，例如，您可以看到哪些方向以及該方向連接到哪個房間。 稍后，您還可以使用有關每個房間的其他信息來擴展字典。 （例如某些房間的特色）

我對此的嘗試：

# Room dictionaries: Each key is a direction and returns the
# room number it connects to. E.g. room1 'forward' goes to room2.
room1 = {'forward': 2, 
         'backward': 3}

room2 = {'forward': 4, 
         'backward': 1,
         'left': 3}

room3 = {'forward': 5, 
         'backward': 1,
         'right': 2}

roomlist = [room1, room2, room3]

def explore_room(roomindex, room):

    # roomindex is the list index (not room number) of the selected room.
    # room is the dictionary describing the selected room.

    question = None
    move_on = None

    print("You are in room %s, how to get out..." % (roomindex+1))
    # Room number added to help you keep track of where you are.

    next_room = True
    while next_room:

        if question == 'exit' or move_on == 'exit':
            break
            # If the user replied 'exit' to either question the loop breaks.
            # Just to avoid an infinite loop.

        question = input().lower()
        # Added the .lower here so we don't have to repeat it in each if.

        # Pick a direction. The response must be valid AND
        # it must be a valid direction for the current room.
        if question == "look forward" and 'forward' in room.keys():
            print("You are looking forward")
            direction = 'forward'
        elif question == 'look backward' and 'backward' in room.keys():
            print("You are looking backward")
            direction = 'backward'
        elif question == 'look left' and 'left' in room.keys():
            print('You are looking to the left')
            direction = 'left'
        # ...
        else:
            print('There is nowhere else to go!')
            continue 
            # Start the loop over again to make the user pick
            # a valid direction.

        # Choose to move on or stay put.
        move_on = input().lower()

        if move_on == 'yes':
            roomindex = room[direction]-1
            # Direction is the dictionary key for the selected direction.
            # Room number is roomindex-1 due to zero-indexing of lists.
            return roomindex
            # Exits explore_room and tells you which room to go to next.

        elif move_on == 'no':
            continue # Ask for a new direction.

        else:
            print('Please answer yes or no')
            continue # Ask for a new direction.

        return None 
        # No room selected. While return None is implicit I included it here for clarity.


roomindex = 0

while True: 
    if roomindex is None:
        # User exited the function without choosing a new room to go to.
        break

    roomindex = explore_room(roomindex, roomlist[roomindex])

您在1號房，如何下車...

向前看

你很期待

是

您在2號房，如何下車...

看左邊

您正在向左看

是

您在3號房，如何下車...

向后看

你回頭看

是

您在1號房，如何下車...

退出無處可去！

>

我承認，是/否不是最受啟發的方法。 無論如何，您可能想將實際問題添加到輸入中，以便播放器可以告訴您對它們的期望。 但總的來說，這似乎可行，您可以輕松地添加和互連任意數量的房間。 所有這些都應該僅使用一種功能。

關於if的鏈接和嵌套：嵌套的if通常是可以的，但是正如furas所說的那樣，它們可能會失去控制並膨脹您的代碼。 但是，根據您的情況，您不需要嵌套的ifs。 如我的示例所示，您可以先問路，然后檢查玩家是否想單獨去那里。 兩者是完全獨立的，因此不需要嵌套語句。

編輯

一個基本的單問題變體，再次受到furas原始答案的啟發：

def explore_room(roomindex, room):

    # roomindex is the list index of the selected room.
    # room is the dictionary describing the selected room.

    question = None

    print("You are in room %s, how to get out..." % (roomindex+1))
    # Roomnumber added to help you keep track of where you are.

    next_room = True
    while next_room:

        if question == 'exit':
            break
            # If the user replied 'exit' to either question the loop breaks.
            # Just to avoid an infinite loop.

        question = input('Choose an action: ').lower()
        # Added the .lower here so we don't have to repeat it each time.

        # Pick a direction. The response must be valid AND
        # it must be a valid direction in this room.
        if question == "look forward" and 'forward' in room.keys():
            print("You are looking forward")
            direction = 'forward'
        elif question == 'look backward' and 'backward' in room.keys():
            print("You are looking backward")
            direction = 'backward'
        elif question == 'look left' and 'left' in room.keys():
            print('You are looking to the left')
            direction = 'left'
        # ...
        elif question == 'move': 
            roomindex = room[direction]-1
            return roomindex
            # Fails if player selects 'move' before picking a direction.
            # It's up to you how you want to handle that scenario.
        else:
            print('Valid actions are "look forward/backward/left/right" and "move". Enter "exit" to quit.')
            continue 
            # Start the loop over again to make the user pick
            # a valid direction.
    return None