我试图让我的代码在屏幕中间显示一个文本,一旦一个正方形向右经过500像素,但它似乎显示我的If条件。我不知道我做错了什么
import pygame
pygame.init()
displaywidth=500
displayheight=500
gameDisplay = pygame.display.set_mode((displaywidth, displayheight))
pygame.display.set_caption("First Game")
red = (0,255,0)
font=pygame.font.Font(None, 20)
#Class
def Message(msg, color):
screen_text=font.render(msg,True,color)
gameDisplay.blit(screen_text,[displaywidth/2,displayheight/2])
win = pygame.display.set_mode((displaywidth, displayheight))
x = 50
y = 50
width = 40
height = 40
vel = 5
x1 = 0
y1 = 0
width1 = 40
height1 = 40
vel2 = 100
vel3=100
x2 = 100
y2 = 100
pygame.draw.rect(win, (0, 255, 0), (x, y, width, height))
run = True
while run:
pygame.time.delay(100)
for event in pygame.event.get():
if event.type == pygame.QUIT:
run = False
keys = pygame.key.get_pressed()
if keys[pygame.K_LEFT]:
x -= vel2
elif keys[pygame.K_RIGHT]:
x += vel2
elif keys[pygame.K_UP]:
y1-=vel3
elif keys[pygame.K_DOWN]:
y1+=vel3
if x > 500:
Message("meow", red)
pygame.display.update()
print("pew")
elif x < 0:
x = 50
elif y1 > 500:
y1 = 450
elif y1 < 0:
y1 = 50
print(x)
print(y)
win.fill((0, 0, 0))
meow = pygame.draw.rect(win, (0, 255, 0), (x, y1, width, height))
pygame.draw.rect(win, (160, 0, 0), (x1, y1, width1, height1))
pygame.display.update()
pygame.quit()
我的打印命令似乎是工作,但我不知道为什么它不显示
问题是您的代码中有2个对
pygame.display.update()
的调用,但在第一个调用之后显示立即被清除:在绘制任何内容之前清除显示,并在主应用程序循环的末尾执行单个
pygame.display.update()
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