如何在时间间隔后显示图像?

Question

我想在用户单击鼠标左键 3 秒后显示图像。 这是我代码的一部分：

pic=pygame.image.load('pic.png')
while True:
for event.type==pygame.MOUSEBUTTONDOWN:
screen.blit(pic,(100,100))

它只显示片刻。我尝试使用for和while循环，但是，它断断续续几秒钟，然后显示闪光灯。

我想我也许可以使用计时器，添加 3s，如下所示：

for event.type==pygame.MOUSEBUTTONDOWN:
#get now time here,and assignment for timeclick
if timeclick+3s>=timenow:  # pseudocode
screen.blit(pic,(100,100))

如何编写此代码段落？还有更好的方法吗？

Answer 1

当用户单击鼠标按钮时启动计时器，然后在主循环中计算经过的时间，如果>= 3，则对图像进行点亮。

import pygame as pg

def main():
screen = pg.display.set_mode((640, 480))
clock = pg.time.Clock()
font = pg.font.Font(None, 40)
img = pg.Surface((100, 100))
img.fill((190, 140, 50))
click_time = 0
passed_time = 0
done = False
while not done:
for event in pg.event.get():
if event.type == pg.QUIT:
done = True
# Start the timer.
elif event.type == pg.MOUSEBUTTONDOWN:
click_time = pg.time.get_ticks()
screen.fill((30, 30, 30))
if click_time != 0:  # If timer has been started.
# Calculate the passed time since the click.
passed_time = (pg.time.get_ticks()-click_time) / 1000
# If 3 seconds have passed, blit the image.
if passed_time >= 3:
screen.blit(img, (50, 70))
txt = font.render(str(passed_time), True, (80, 150, 200))
screen.blit(txt, (50, 20))
pg.display.flip()
clock.tick(30)

if __name__ == '__main__':
pg.init()
main()
pg.quit()

Answer 2

您必须在主应用程序循环中绘制图像。使用pygame.time.get_ticks()返回自调用pygame.init()以来的毫秒数。当MOUSEBUTTONDOWN事件发生时，计算必须显示图像的时间点。显示当前时间大于计算时间点后的图像：

import pygame
pygame.init()
screen= pygame.display.set_mode((800, 600))
#pic = pygame.image.load('pic.png')
pic = pygame.Surface((100, 100))
pic.fill((255, 255, 255))
pic_time = 0
run = True
while run:
current_time = pygame.time.get_ticks()
for event in pygame.event.get():
if event.type == pygame.QUIT:
run = False
if event.type == pygame.MOUSEBUTTONDOWN:
pic_time = current_time + 3000 # 3000 milliseconds == 3 seconds
screen.fill(0)
if pic_time > 0 and current_time >= image_time:
screen.blit(pic,(100,100))
pygame.display.flip()
pygame.quit()