我使用的是python2.7。重新绘制基于字符串的级别时遇到问题。第一帧画得很好,但之后,就再也画不出来了。我一直在想办法解决这个问题。以下是整个源代码:
import pygame
from pygame.locals import *
xScan = 0
yScan = 0
level = "WWWWWWWW" \
"WSSSSSSW" \
"WSSSSSSW" \
"WSSSSSSW" \
"WSSSSSSW" \
"WWWWWWWW"
class Player():
def __init__(self, color, x, y, width, height, outline_size):
self.color = color
self.x = x
self.y = y
self.width = width
self.height = height
self.outline_size = outline_size
def draw(self):
pygame.draw.rect(screen, self.color, [self.x * 80, self.y * 80, self.width * 80, self.height * 80], self.outline_size)
def draw_level():
for tile in level:
global xScan, yScan
if tile == "W":
pygame.draw.rect(screen, [0, 0, 0], [xScan * 80, yScan * 80, 80, 80], 0)
elif tile == "S":
pygame.draw.rect(screen, [120, 120, 120], [xScan * 80, yScan * 80, 80, 80], 1)
if xScan >= 0 and xScan < 7:
xScan += 1
elif xScan == 7:
xScan = 0
yScan += 1
elif yScan == 5 and xScan == 7:
xScan = 0
yScan = 0
def update():
screen.fill([255, 255, 255])
draw_level()
player.draw()
pygame.display.flip()
pygame.init()
screen = pygame.display.set_mode([640, 480])
pygame.display.set_caption("Squares n' Tiles!")
player = Player([0, 100, 255], 1, 1, 1, 1, 0)
running = True
while running:
clock = pygame.time.Clock()
clock.tick(30)
for event in pygame.event.get():
if event.type == pygame.KEYDOWN:
if event.key == K_ESCAPE:
running = False
if event.key == K_UP and player.y * 80 >= 160:
player.y -= 1
if event.key == K_DOWN and player.y * 80 < 320:
player.y += 1
if event.key == K_LEFT and player.x * 80 >= 160:
player.x -= 1
if event.key == K_RIGHT and player.x * 80 < 480:
player.x += 1
if event.key == K_c or event.key == K_r:
player.x = 1
player.y = 1
if event.type == pygame.QUIT:
running = False
update()
pygame.quit()
我认为问题在于这部分代码:
^{pr2}$我试着在前一部分的后半部分弄乱数字。以下是一个主要解决问题的变体(尽管其格式不是很一致):
if xScan >= 7:
xScan = 0
yScan += 1
elif yScan == 6:
xScan = 1
yScan = 0
else:
xScan += 1
上面代码的唯一问题是左上角的平铺无法绘制。我如何改变我的代码使其正常工作?在每一帧上,我希望屏幕上填充白色,重新绘制关卡,重新绘制玩家,并翻转屏幕。提前谢谢你的帮助。我也希望能对我的代码结构提出建议。在
问题是这个块没有被执行:
这是因为在
level
字符串中的每个字符上循环,所以达到的最大值是xScan=6,yScan=5。在相反,在进入draw level函数时只需初始化}
^{pr2}$xScan
和{我建议存储几行而不是一行长的字符:
然后可以在地图上进行迭代,如下所示:
^{4}$enumerate
获取一个list(或iterable),并为列表中的每个项返回成对的(index,item),例如:相关问题 更多 >
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