Question

我在Python中制作一个简单的基于文本的RPG。目前我对大多数房间有两种方法，一种用于第一次进入，一种用于返回。有没有办法可以确保他们没有其他方法之前没有进过那个房间？

例如，如果我有一个名为tomb()的方法，我创建另一个名为tombAlready()的方法，该方法包含相同的代码，除了房间的介绍文本。

所以，如果我有

slow_type("\n\nTomb\n\n")
  slow_type("There is an altar in the middle of the room, with passages leading down and west.")
  choice = None
  while choice == None:
    userInput = input("\n>")
    if checkDirection(userInput) == False:
      while checkDirection == False:
        userInput = input("\n>")
        checkDirection(userInput)
    userInput = userInput.lower().strip()
    if userInput == "d":
      catacombs()
    elif userInput == "n":
      altar()
    elif userInput == "w":
      throneroom()
    else:
      slow_type("You cannot perform this action.")

然后tombAlready()除了slow_type("There is an altar in the middle of the room, with passages leading down and west.")

之外会有相同的代码

Answer 1

您想要的是与功能相关联的状态。使用带有方法的对象：

class Room:
    def __init__(self, description):
        self._description = description
        self._visited = False

    def visit(self):
        if not self._visited:
            print(self._description)
            self._visited = True

然后你可以为每个房间设置一个Room对象：

catacombs = Room('There is a low, arched passageway. You have to stoop.')
tomb = Room('There is an altar in the middle of the room, with passages leading down and west.')
throneroom = Room('There is a large chair. It looks inviting.')

您可以访问一个房间两次，但它只打印一次描述：

>>> catacombs.visit()
There is a low, arched passageway. You have to stoop.

>>> catacombs.visit()

如何检查用户是否已经通过方法

1 个答案: