Python 如何在不做任何事情的情况下改变矩形的颜色？

我想画一个矩形，它可以在用户不做任何事情的情况下从黑色变为红色。下面的代码应该填充屏幕，画一个黑色矩形，然后删除它，再画一个红色矩形，然后继续循环。但我只得到一个红色矩形。我做错了什么

import pygame

# Define some colors
BLACK = (0, 0, 0)
WHITE = (255, 255, 255)
GREEN = (0, 255, 0)
RED = (255, 0, 0)

pygame.init()

# Set the width and height of the screen [width, height]
size = (700, 500)
screen = pygame.display.set_mode(size)

pygame.display.set_caption("My Game")

# Loop until the user clicks the close button.
done = False

# Used to manage how fast the screen updates

clock = pygame.time.Clock()


# -------- Main Program Loop -----------
while not done:
    # --- Main event loop
    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            done = True
    screen.fill(WHITE)
    pygame.draw.rect(screen, BLACK, [300, 200, 100, 100],0)
    screen.fill(WHITE)
    pygame.draw.rect(screen, RED, [300, 200, 100, 100],0)

    pygame.display.flip()

    clock.tick(60)


pygame.quit()

---

您可以在

rect（…）

和..中使用带颜色的变量

。。。将自己的事件与

pygame.time一起使用。设置\u timer（）

可更改此变量中的颜色

import pygame

BLACK = (0, 0, 0)
WHITE = (255, 255, 255)
GREEN = (0, 255, 0)
RED = (255, 0, 0)

size = (700, 500)

# --- start ---

pygame.init()
screen = pygame.display.set_mode(size)

# start color
color = RED

# define own event type
CHANGE_COLOR = pygame.USEREVENT + 1
# create event every 250ms
pygame.time.set_timer(CHANGE_COLOR, 250) # 250ms

# --- mainloop ---

done = False
clock = pygame.time.Clock()

while not done:

    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            done = True

        # check if get event
        if event.type == CHANGE_COLOR:
            # change color
            if color == RED:
                color = BLACK
            else:
                color = RED

    screen.fill(WHITE)
    pygame.draw.rect(screen, color, [300, 200, 100, 100],0)
    pygame.display.flip()

    clock.tick(60)

pygame.quit()

。。。或者使用

pygame.time.get_ticks（）

获取当前时间并检查是否是更改颜色的时间

import pygame

BLACK = (0, 0, 0)
WHITE = (255, 255, 255)
GREEN = (0, 255, 0)
RED = (255, 0, 0)
size = (700, 500)

# --- start ---

pygame.init()
screen = pygame.display.set_mode(size)

# start color    
color = RED

# get current time
current_time = pygame.time.get_ticks()

# first change after 250 ms
change_color_time = current_time + 250

# --- mainloop ---

done = False
clock = pygame.time.Clock()

while not done:

    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            done = True

    # get current time 
    current_time = pygame.time.get_ticks()

    # check if it is time to change color
    if current_time >= change_color_time:
        # set new time to change color again
        change_color_time = current_time + 250
        # change color
        if color == RED:
            color = BLACK
        else:
            color = RED                    

    screen.fill(WHITE)
    pygame.draw.rect(screen, color, [300, 200, 100, 100],0)
    pygame.display.flip()

    clock.tick(60)

pygame.quit()

---

如果你1<代码>导入时间和2。在绘制黑色矩形以延迟红色矩形后，添加

time.sleep（3）

仅使用一次，并在

rect（…，当前颜色，…）

中使用变量，您将在循环中更改该变量。并使用

pygame.time.Clock（）.tick（）

来减慢速度，或者使用计时器（或事件）来控制何时更改颜色。