Pygame：当精灵跑进它的轨道时，如何“输掉”游戏

Question

我创建了一个类似蛇的游戏，用户移动精灵，精灵会留下痕迹。如果用户遇到他创建的轨迹，我希望游戏结束，玩家输。你知道吗

我的一个想法是跟踪精灵过去的位置（可能在列表中），然后创建一个“如果”语句，这将导致游戏失败（然而，我有点不清楚如何做到这一点）。你知道吗

我收到了这个问题的答案，这个答案编码如下：

"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 == pastPositions[row][0] and player.rect.y == >pastPositions[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])"

这看起来应该可以工作，但是当我尝试运行代码（在Python交互式中）时，会收到一条错误消息：“line 86，in if播放器.rect.x==ast位置[行][0]和播放器.rect.y==pastPositions[row][1]：索引器错误：列表索引超出范围“–Gal 2天前

你建议我把范围改成什么，这样就不会发生这种情况了？我试着把它设置为pygame窗口的宽度和高度。你知道吗

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

我想你可以这样声明一个二维列表：

pastPositions = [[400, 300]]

然后每次球员的位置移动时，检查列表：

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

如果玩家还没去过那里，那么把这个位置添加到列表中。你知道吗

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

你在找这样的东西吗？你知道吗