Question

我创建了一个类似蛇的游戏，用户在其中移动精灵，精灵留下轨迹。如果用户遇到了他创建的足迹，我希望游戏结束，而玩家则输了。

我的一个想法是以某种方式跟踪精灵的过去位置（也许在列表中），然后创建一个“ if”语句，这将导致游戏失败（但是，我不清楚如何为此）。

我收到了针对以下列表的编码问题的答案：

”“我认为您可以像这样声明一个二维列表：    pastPositions = [[400，300]]   然后，每次玩家位置移动时，请检查以下列表：
  对于pastPositions中的行：       如果（player.rect.x == pastPositions [row] [0]和player.rect.y ==> pastPositions [row] [1]）：            完成=正确＃游戏结束   如果玩家还没来过，那么将该位置添加到列表中。   pastPositions.append（[player.rect.x，player.rect.y]）“

这看起来应该可行，但是当我尝试运行代码（在Python Interactive中）时，出现一条错误消息，内容为：“第86行，如果player.rect.x == astPositions [row] [ 0]和player.rect.y == pastPositions [row] [1]：IndexError：列表索引超出范围” – Gal 2天前

您建议我将范围更改为什么，以免发生这种情况？我尝试将其设置为pygame窗口的宽度和高度。

import pygame

BLACK = (0, 0, 0)
WHITE = (255, 255, 255)

class Player(pygame.sprite.Sprite):

   def __init__(self, x, y):
       super().__init__()

       self.image = pygame.Surface([15, 15])
       self.image.fill(WHITE)

       self.rect = self.image.get_rect()
       self.rect.x = x
       self.rect.y = y

       self.change_x = 0
       self.change_y = 0


   def changespeed(self, x, y):
       self.change_x += x
       self.change_y += y


   def update(self):
       self.rect.x += self.change_x
       self.rect.y += self.change_y

pygame.init()

screen = pygame.display.set_mode([800, 600])

pygame.display.set_caption('The Etch-a-Sketch Game')

myfont = pygame.font.SysFont('Times', 20)
textsurface = myfont.render('This is the Etch-a-Sketch Game', False, (255, 255, 255))
screen.blit(textsurface,(0,0))

myfont = pygame.font.SysFont('Times', 15)
textsurface = myfont.render('Feel free to draw, but if you cross your own    path, you will die.', False, (255, 255, 255))
screen.blit(textsurface,(0,20))


player = Player(400, 300)
all_sprites_list = pygame.sprite.Group()
all_sprites_list.add(player)

clock = pygame.time.Clock()
done = False

while not done:

    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            done = True

        elif event.type == pygame.KEYDOWN:
            if event.key == pygame.K_LEFT:
                player.changespeed(-3, 0)
            elif event.key == pygame.K_RIGHT:
                player.changespeed(3, 0)
            elif event.key == pygame.K_UP:
                player.changespeed(0, -3)
            elif event.key == pygame.K_DOWN:
                player.changespeed(0, 3)

        elif event.type == pygame.KEYUP:
            if event.key == pygame.K_LEFT:
                player.changespeed(3, 0)
            elif event.key == pygame.K_RIGHT:
                player.changespeed(-3, 0)
            elif event.key == pygame.K_UP:
                player.changespeed(0, 3)
            elif event.key == pygame.K_DOWN:
                player.changespeed(0, -3)

    player.update()

    all_sprites_list.draw(screen)

    pygame.display.flip()
    clock.tick(75)

pygame.quit ()

Answer 1

I think you could declare a two dimensional list like this:

pastPositions = [[400, 300]]

Then every time the player's position moves, check the list:

for row in pastPositions: if (player.rect.x == row[0] and player.rect.y == row[1]): done = true # game over

If the player hasn't been there yet, then add that position to the list.

pastPositions.append([player.rect.x, player.rect.y])

Are you looking for something like that?

Pygame：精灵运行时如何“丢失”游戏

1 个答案: