今天我在pygame中制作了一个小平台游戏。我从一个非常漂亮和有用的教程中使用了很多代码，因为我不确切知道如何做到这一点。
到目前为止，通过处理屏幕的32x32部分并根据地图文本部分中的字符为其分配块类型来制作地图。到目前为止，我有一个平台块，一个边界块和一个出口块，我想最终让游戏加载到一个新的水平。
我的问题是：如果我“超越水平”并希望让游戏加载新地图，我该如何去做呢？我试过从组中删除精灵，但我没有运气。我已经尝试了一段时间，我有点气馁。
以下是代码：
import pygame
import math
from pygame import *
DISPLAY = (800, 640)
DEPTH = 32
FLAGS = 0
def main():
pygame.init()
screen = display.set_mode(DISPLAY, FLAGS, DEPTH)
display.set_caption("Square2")
timer = time.Clock()
up = down = left = right = False
bg = Surface((32,32))
bg.convert()
bg.fill(Color("#000000"))
entities = pygame.sprite.Group()
player = Player(32,32)
platforms = []
x = y = 0
level = [
"PPPPPPPPPPPPPPPPPPPPPPPPP",
"P P",
"P P",
"P P",
"P P",
"P P",
"P P",
"P P",
"P P",
"P P",
"P P",
"P P",
"P pp P",
"P P",
"P P",
"P pppppp P",
"P P",
"P pppppp P",
"P P",
"PPPPPPPPPPPPPPPPPPPPPPPPP",
]
#builds the level
for row in level:
for col in row:
if col =="p":
p = Platform(x, y)
platforms.append(p)
entities.add(p)
if col == "E":
e = ExitBlock(x, y)
platforms.append(e)
entities.add(e)
if col == "P":
P = Border(x, y)
platforms.append(P)
entities.add(P)
x += 32
y += 32
x = 0
entities.add(player)
while 1:
timer.tick(60)
for e in pygame.event.get():
if e.type == QUIT:
pygame.quit()
sys.exit()
#raise SystemExit, "ESCAPE"
if e.type == KEYDOWN and e.key == K_ESCAPE:
pygame.quit()
sys.exit()
#raise SystemExit, "ESCAPE"
if e.type == KEYDOWN and e.key == K_UP:
up = True
if e.type == KEYDOWN and e.key == K_DOWN:
down = True
if e.type == KEYDOWN and e.key == K_RIGHT:
right = True
if e.type == KEYDOWN and e.key == K_LEFT:
left = True
if e.type == KEYUP and e.key == K_UP:
up = False
if e.type == KEYUP and e.key == K_DOWN:
down = False
if e.type == KEYUP and e.key == K_RIGHT:
right = False
if e.type == KEYUP and e.key == K_LEFT:
left = False
#draw bg
for y in range(20):
for x in range(25):
screen.blit(bg, (x * 32, y * 32))
#update player, draw everything else
player.update(up, down, left, right, platforms)
entities.draw(screen)
pygame.display.flip()
class Entity(pygame.sprite.Sprite):
def __init__(self):
pygame.sprite.Sprite.__init__(self)
class Player(Entity):
def __init__(self, x, y):
Entity.__init__(self)
self.xvel = 0
self.yvel = 0
self.onGround = False
self.image = Surface((32, 32))
self.image.convert()
self.image.fill(Color("#FF0000"))
self.rect = Rect(x, y, 32, 32)
self.reset(150, 400)
def reset(self,x,y):
self.x = x
self.y = y
self.rect.center = (self.x, self.y)
def update(self, up, down, left, right, platforms):
if up:
#only jump if on the ground
if self.onGround:
self.yvel = -8
if down:
pass
if left:
self.xvel = -5
if right:
self.xvel = 5
if not self.onGround:
#only accelerate wit gravity if in the air
self.yvel +=0.3
#max falling sapeed
if self.yvel > 30:
self.yvel = 30
if not (left or right):
self.xvel = 0
# increment in x direction
self.rect.left+= self.xvel
# x-axis collisions
self.collide(self.xvel, 0 , platforms)
# increment in y direction
self.rect.top +=self.yvel
# assuming we're in the air
self.onGround = False;
# y-axis collisions
self.collide(0, self.yvel, platforms)
def collide(self, xvel, yvel, platforms):
for p in platforms:
if sprite.collide_rect(self, p):
if isinstance(p, ExitBlock):
event.post(event.Event(QUIT))
if xvel > 0:
self.rect.right = p.rect.left
if xvel < 0:
self.rect.left = p.rect.right
if yvel > 0:
self.rect.bottom = p.rect.top
self.onGround = True
self.yvel = 0
if yvel < 0:
self.rect.top = p.rect.bottom
self.yvel = 0
# Player.rect.top = Player.rect.top
class Platform(Entity):
def __init__(self, x, y):
Entity.__init__(self)
self.image = Surface((32, 32))
self.image.convert()
self.image.fill(Color("#DDDDDD"))
self.rect = Rect(x, y, 32, 32)
def update(self):
pass
class Border(Entity):
def __init__(self, x, y):
Entity.__init__(self)
self.image = Surface((32, 32))
self.image.convert()
self.image.fill((0,0,0))
self.rect = Rect(x, y, 32, 32)
def update(self):
pass
class ExitBlock(Platform):
def __init__(self, x, y):
Platform.__init__(self, x, y)
self.image.fill(Color("#0033FF"))
main()
提前感谢您的帮助！
答案 0 :(得分：0)
创建一个算法来随机创建一个可行的级别并输出一个级别的可读字符串。自动将其与边框对齐，制作一些随机平台，以及距离播放器最远的平台上的出口。您甚至可以计算是否可以从一个平台跳到下一个平台，您可能应该这样做。要加载您的关卡，只需重新调用main（）即可。此外，实体看起来它只是精灵的包装，你可能想保留它，但它似乎只是占用空间。因为算法可能是整个程序中最大的一部分，所以我不会写一个例子，但是如果你想要一种方法来加载一个新的水平或一个地方开始算法，这应该有效。