Question

经过数小时的搜索，我仍然无法弄清为什么只有最近生成的圆才受到碰撞检测的影响。我注释掉了有问题的代码。我尝试了精灵，这也许是答案，但我仍然得到相同的结果。

import pygame,random
pygame.init()
width,height,radius = 1280,720,20
class Ball(): 
    def __init__(self):  
        self.x = 0
        self.y = 0
        self.vx = 0
        self.vy = 0  
def make_ball():
    ball = Ball()
    ball.x = random.randrange(radius, width - radius)
    ball.y = random.randrange(radius, 100)
    ball.vx = random.randint(1,2)
    ball.vy = 0
    return ball
def main():
    rect_x = 60    
    display = pygame.display.set_mode((width,height))
    pygame.display.set_caption("BOUNCE")
    running = True
    ball_list = []
    ball = make_ball()
    ball_list.append(ball)
    while running:  
        for event in pygame.event.get():
            if event.type == pygame.QUIT:
                running = False
            elif event.type == pygame.KEYDOWN:
                if event.key == pygame.K_LEFT or event.key == pygame.K_RIGHT:
                    ball = make_ball()
                    ball_list.append(ball)
        for ball in ball_list: 
            ball.x += ball.vx
            ball.vy += 0.02
            ball.y += ball.vy
            if ball.y >= height - radius:
                ball.vy *= -1
            if ball.x >= width - radius or ball.x <= radius:
                ball.vx *= -1    
        display.fill((0,0,0))   
        for ball in ball_list:
            random_color = (random.randint(1,255),random.randint(1,255),random.randint(1,255))
            circle = pygame.draw.circle(display,random_color,(int(ball.x), int(ball.y)),radius)   
        rectangle = pygame.draw.rect(display,(255,255,255),(int(rect_x),660,60,60)) 
        if event.type == pygame.KEYDOWN:
            if event.key == pygame.K_LEFT and rect_x > 0:
                rect_x -= 2
            if event.key == pygame.K_RIGHT and rect_x < width - 60:
                rect_x += 2
        '''if pygame.Rect(circle).colliderect(rectangle) == True:   ###THIS IS THE BAD CODE!
            print('Your Score:',pygame.time.get_ticks())
            running = False'''
        text = pygame.font.Font(None,120).render(str(pygame.time.get_ticks()),True,(255,255,255))
        display.blit(text,(50,50)) 
        pygame.display.flip()
    pygame.quit()
if __name__ == "__main__":
    main()

Answer 1

缩进和代码组织是实现此目的的关键。令人讨厌的部分是（删除了评论）：

for ball in ball_list:
    random_color = (random.randint(1,255),random.randint(1,255),random.randint(1,255))
    circle = pygame.draw.circle(display,random_color,(int(ball.x), int(ball.y)),radius)   
rectangle = pygame.draw.rect(display,(255,255,255),(int(rect_x),660,60,60)) 
if event.type == pygame.KEYDOWN:
    if event.key == pygame.K_LEFT and rect_x > 0:
        rect_x -= 2
    if event.key == pygame.K_RIGHT and rect_x < width - 60:
        rect_x += 2
if pygame.Rect(circle).colliderect(rectangle) == True:
    print('Your Score:',pygame.time.get_ticks())
    running = False

您拥有所有正确的零件，但是关闭它们的顺序以及缩进：

    for ball in ball_list:
        random_color = (random.randint(1,255),random.randint(1,255),random.randint(1,255))
        circle = pygame.draw.circle(display,random_color,(int(ball.x), int(ball.y)),radius)
        rectangle = pygame.draw.rect(display,(255,255,255),(int(rect_x),660,60,60))
        if pygame.Rect(circle).colliderect(rectangle):
            print('Your Score:',pygame.time.get_ticks())
            running = False

现在它将遍历列表中的每个球，并检查每个球是否发生碰撞。注意colliderect if语句缩进for循环中。还要注意，我从所有中间删除了KEYDOWN检查

谈到这一点，我建议使用：

pressed = pygame.key.get_pressed()
if pressed[pygame.K_LEFT] and rect_x > 0:
    rect_x -= 2
if pressed[pygame.K_RIGHT] and rect_x < width - 60:
    rect_x += 2

for ball in ball_list:
    # for loop from above

而不是您拥有的。当您要允许按住一个键时，这是最合适的。 pygame.key.get_pressed（）始终获取所有键的状态，而不仅仅是事件发生时

为什么此冲突检测不适用于先前创建的对象？

1 个答案: