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Question

我在自己想出的程序构想中练习类和继承。基本上，我制作了一个可以玩两种模式的街机游戏菜单模拟器，即单人游戏和多人游戏。每次输入一个选项1或2时，菜单都会显示几次，然后继续接受输入，我只希望菜单显示一次。这是我的代码：

    # Suppose you are at an arcade and you and your friend want to play a multiplayer game that requires UI.
# Make the game ask for the users name and age to see if they can play, make the program so that it can add a friend.
# If any of the players are under the age of 18, they are not allowed to  play, otherwise proceed.
# **EXTRA CREDIT** --> Add a functionality which adds the players to a list until the list reaches 4 players, then stop adding to the list.

# arcade_game.py
import sys


# give the user a greeting
import self as self

lst = []


class menu:
    def __init__(self, ready):
        self.ready = ready

    #display menu
    @classmethod
    def display_menu(self):
        print("Pick from one of the choices below, type in the corressponding number")
        print("1. single player \n"
              "2. Multiplayer")
        choice = int(input("Enter your choice here: "))
        return choice

    # ready or not function to see if the user is ready to play
    def ready_or_not(self):
        # see if user types 1 or 2 with try & except
        try:
            # ask user if they are ready
            self.ready = int(input("Are you ready to play? Type 1 for yes, 2 for no"))
            self.display_menu()
        except ValueError:
            print("You did not type 1 or 2, please try again!")


# add players class
class player(menu):
    # add a default player to __init__(), **(since there has to be at least one player)**
    def __init__(self, ready, player1):
        super().__init__(ready)
        self.player1 = player1

    # single player method
    def set_name(self):
        self.player1 = input("Enter your name for single player mode")
        print("Lets play! ", self.player1)

    # multiplayer method
    def set_names(self):
        try:
            self.player1 = input("Enter your name to begin")
            lst.append(self.player1)
            # add another player to continue
            while len(lst) <= 4:
                add = input("Add player here: ")
                lst.append(add)
                if len(lst) == 4:
                    print("Player limit reached!")
                    break;
        except ValueError:
            print("You didnt enter valid input, please try again")

    # get the names of the players only if 1 is picked from display_menu() above, including player1
    def check_choice(self):
        if self.display_menu() == 1:
            self.set_name()
        elif self.display_menu() == 2:
            self.set_names()
        else:
            print("Exiting....")
            print("Goodbye!")
            sys.exit(0)


m = menu("yes")
m.ready_or_not()
p = player("yes", "test")
p.check_choice()

Answer 1

ready_or_not呼叫self.display_menu()：

def ready_or_not(self):
    # see if user types 1 or 2 with try & except
    try:
        # ask user if they are ready
        self.ready = int(input("Are you ready to play? Type 1 for yes, 2 for no"))
        self.display_menu()
    except ValueError:
        print("You did not type 1 or 2, please try again!")

check_choice还会调用self.display_menu()至少一次，如果您第一次键入除1以外的其他值，则两次调用两次：

def check_choice(self):
    if self.display_menu() == 1:
        self.set_name()
    elif self.display_menu() == 2:
        self.set_names()
    else:
        print("Exiting....")
        print("Goodbye!")
        sys.exit(0)

您的顶级代码在一个菜单实例上调用ready_or_not()：

m = menu("yes")
m.ready_or_not()

…和check_choice()在另一个：

p = player("yes", "test")
p.check_choice()

因此，您的程序将菜单显示两次，如果您键入除1以外的任何内容，则将显示第三次。

如果您不希望菜单显示两次或三次，请不要调用方法两次或三次。

如果您只想显示菜单一次并记住选择，而不是显示两次或三遍，则需要使用在self.ready中创建的ready_or_not属性，而不是调用再次使用该方法。

但是，这仍然无法按原样工作，因为您的类设计很奇怪。您已经创建了两个单独的实例m和p，每个实例都有自己的独立属性。我不确定为什么player首先从menu继承（或者为什么display_menu是@classmethod，或者为什么它调用其参数self而不是为什么）而不是cls（如果是一个，以及其他各种东西），但是，鉴于设计中的player是menu，您可能只需要一个player实例，就像这样：

p = player("yes", "test")
p.ready_or_not()
p.check_choice()

然后，您可以像这样更改check_choice：

def check_choice(self):
    if self.choice == 1:
        self.set_name()
    elif self.choice == 2:
        self.set_names()
    else:
        print("Exiting....")
        print("Goodbye!")
        sys.exit(0)

Answer 2

花点时间找出答案，但是看来display_menu()的通话ready_or_not()已经结束，因此您需要将display_menu()从准备就绪状态中删除，否则就不这样 >

   # ready or not function to see if the user is ready to play
    def ready_or_not(self):
        # see if user types 1 or 2 with try & except
        try:
            # ask user if they are ready
            self.ready = int(input("Are you ready to play? Type 1 for yes, 2 for no"))
#            self.display_menu()
        except ValueError:
            print("You did not type 1 or 2, please try again!")

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参加聚会似乎很晚

输入输入后，我的显示菜单会重复多次

2 个答案:

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