Question

我为游戏编写了一个代码，您在其中滑动15个图块以按顺序对其进行设置。在游戏结束时，我希望屏幕仅等待5秒钟才能关闭（也希望稍后再提示）。

游戏结束时，它首先暂停，然后显示最后一个动作，然后关闭。显示屏是否有滞后。如何使它完成显示，然后等待。我尝试阅读类似的帖子，但听不懂。

这是相关的代码。

>   def run_game(self):
        """Start the main loop for the game"""
        self._shuffle_grid()
        self._show_board()
        game_over = False
        while True:
            change = self._check_events()
            if change:
                self._show_board()
                game_over = self.end_of_game()
            if game_over:
                pygame.time.wait(5000)
                sys.exit()


    def _check_events(self):
        # Watch for keyboard and mouse events
        change = False
        for event in pygame.event.get():
            if event.type == pygame.QUIT:
                sys.exit()
            elif event.type == pygame.KEYDOWN:
                if event.key == pygame.K_UP:
                    key = "U"
                elif event.key == pygame.K_DOWN:
                    key = "D"
                elif event.key == pygame.K_LEFT:
                    key = "L"
                elif event.key == pygame.K_RIGHT:
                    key = "R"
                if key == "U" or "D" or "L" or "R":
                    change = self.key_check(key)
        return(change)

Answer 1

我实际上尝试了这个并且奏效了。我仍然不知道如何显示消息，然后等待5秒钟。正在努力。

    def celebrate(self):
        """ does end of game thing"""
        game_clock = pygame.time.Clock()
        time_passed = pygame.time.get_ticks()
        self._show_board()
        while True:
            for event in pygame.event.get():
                if event.type == pygame.QUIT or (event.type == pygame.KEYDOWN and event.key == pygame.K_ESCAPE):
                    sys.exit()
            if pygame.time.get_ticks() > time_passed + 5000:
                sys.exit()


    def run_game(self):
        """Start the main loop for the game"""
        self._shuffle_grid()
        self._show_board()
        game_over = False
        while True:
            change = self._check_events()
            if change:
                game_over = self.end_of_game()
                if not game_over:               
                    self._show_board()
                else:
                    self.celebrate()

为什么pygame.time.wait在显示更新之前启动？

1 个答案: