Question

你好我对使用pygame编程很新，我有这个代码：

import sys, pygame

pygame.init()
screen = pygame.display.set_mode((800, 368))
background = pygame.image.load("background.png")
background.convert()
screen.blit(background, (0, 0))

speed = [1, 1]
width = 800
height = 368

ball = pygame.image.load("ball.bmp")
ballrect = ball.get_rect()
player1 = pygame.image.load("player1.png")
player1rect = player1.get_rect()

while 1:
    for event in pygame.event.get():
        if event.type == pygame.QUIT: 
            sys.exit()              

    ballrect = ballrect.move(speed)
    if ballrect.left < 0 or ballrect.right > width:
        speed[0] = -speed[0]
    if ballrect.top < 0 or ballrect.bottom > height:
        speed[1] = -speed[1]

    screen.blit(ball, ballrect)
    screen.blit(player1, player1rect)
    pygame.display.update()

但是当我跑它时会出现太多球，它应该只是一个球。并且有越来越多的球。

Answer 1

不是pygame专家，我认为你需要做类似

的事情

new_ball_rect = ballrect.move(speed)
...
screen.blit(background, ballrect, ballrect) #redraw background over old ball location
screen.blit(ball, new_ball_rect)

在旧球图像上重新绘制背景。

Answer 2

Fredrik Håård关于重新绘制背景是正确的。

screen充当图像，当您致电screen.blit()时，您将绘制该图像的一部分。当while 1:循环重复时，您的图像上已经有一个或多个球的副本，而您当前的代码只是将另一个球的图像绘制到screen上。

您可以在移动球之前绘制背景，从而避免添加新变量来保持前一个球的位置。

...
while 1:
for event in pygame.event.get():
    if event.type == pygame.QUIT:
        sys.exit()

    screen.blit(background, ballrect, ballrect) #redraw background over old ball location
    ballrect = ballrect.move(speed)
    if ballrect.left < 0 or ballrect.right > width:
        speed[0] = -speed[0]
    if ballrect.top < 0 or ballrect.bottom > height:
        speed[1] = -speed[1]

    screen.blit(ball, ballrect)
    screen.blit(player1, player1rect)
    pygame.display.update()

或者，您可以每次重建整个screen，方法是将上面的screen.blit(background, ballrect, ballrect)替换为screen.blit(background, (0, 0))，这样可以确保背景对每个“框架”都正确，但速度要慢得多

我首选的方法是注意您可以传递pygame.display.update()一个矩形列表，这样可以更快地恢复呼叫。在进入while循环之前，请确保执行初始全屏显示更新：

...
pygame.display.update()

while 1:
    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            sys.exit()

    old_ballrect = ballrect
    screen.blit(background, old_ballrect, old_ballrect) #redraw background over old ball location
    ballrect = ballrect.move(speed)
    if ballrect.left < 0 or ballrect.right > width:
        speed[0] = -speed[0]
    if ballrect.top < 0 or ballrect.bottom > height:
        speed[1] = -speed[1]

    screen.blit(ball, ballrect)
    screen.blit(player1, player1rect)
    pygame.display.update([old_ballrect, ballrect])

为什么有超过1个球，我该如何解决？

2 个答案: