Pygame，在屏幕上移动一个方块，但不能擦除之前的移动

Question

这是我在 Whosebug 的第一个 post，希望你们能帮助新手程序员。 Pygame Python 简单的问。

我想在屏幕上移动一个方块，但无法擦除之前的移动。

import pygame

pygame.init()
screen = pygame.display.set_mode((400,300))
pygame.display.set_caption("shield hacking")
JogoAtivo = True
GAME_BEGIN = False
# Speed in pixels per frame
x_speed = 0
y_speed = 0
cordX = 10
cordY = 100


def desenha():
    quadrado = pygame.Rect(cordX, cordY ,50, 52)

    pygame.draw.rect(screen, (255, 0, 0), quadrado)
    pygame.display.flip()



while JogoAtivo:
    for evento in pygame.event.get():
        print(evento);
    #verifica se o evento que veio eh para fechar a janela
        if evento.type == pygame.QUIT:
               JogoAtivo = False
               pygame.quit();
        if evento.type == pygame.KEYDOWN:     
            if evento.key == pygame.K_SPACE:
                   print('GAME BEGIN')
                   desenha()
                   GAME_BEGIN = True;
            if evento.key == pygame.K_LEFT and GAME_BEGIN:   
                   speedX=-3
                   cordX+=speedX
                   desenha()


            if evento.key == pygame.K_RIGHT and GAME_BEGIN:
                   speedX=3
                   cordX+=speedX
                   desenha()

Answer 1

你必须在上一张图片上画画。最简单的方法是在游戏循环开始时用背景色填充屏幕，如下所示：

screen.fill((0, 0, 0))

Answer 2

您也可以在正方形到位之前从以前的表面绘制所有内容，然后始终在不同的位置绘制正方形。然后每次绘制新的表面到屏幕上。

但不是很好的策略，它可能会很慢 and/or 导致闪烁。

例如：

# Draw anything to the display surface named screen, then do:
# By display surface I mean the surface returned by pygame.display.set_mode()
orig = screen.copy()
# Sometimes it will not be filled with a image from screen that you have drawn before, so you draw
# all to orig surface
# Then you make a surface that will be shown on the screen and do:
s = pygame.surface.Surface()
s.blit(orig, 0, 0)
# Then you draw the rectangle or whatever on s, and then:
screen.blit(s, 0, 0)
pygame.display.flip()
# and you keep orig as a starting point for every move

如果屏幕上同时有许多可移动对象，这很好，或者如果您只刷新受限区域，则按照其他答案中的建议重新绘制最后一张图像。

如果直接绘制到显示表面，闪烁会更大，有时，尤其是当显示是硬件加速时，您会遇到问题。尤其是在手机等非桌面设备上。绘制和添加新的通常直接在显示表面上工作正常，并且是解决您的问题的正确方法。

我知道我听起来很矛盾，但有些东西您必须亲自看看。通过试验获得经验。

Answer 3

我正在做类似的事情，希望这对您有所帮助

import pygame

clock=pygame.time.Clock() #the frames per second BIF in pygame
pygame.init()
FPS=30
display_width=800
display_height=600
white=(255,255,255)
black=(0,0,0)
block_size=10
gameExit=False
lead_x = display_width / 2
lead_y = display_height / 2

lead_x_change = 0  # 0 beacuse we will not be changing the position in the beginning
lead_y_change = 0

gameDisplay=pygame.display.set_mode((display_width,display_height))

while not gameExit: #game loop

    for event in pygame.event.get():  #event handling BIF in pygame EVENT LOOP
        #print(event) # prints out the position of the mouse and the buttons pressed when in game window
        if event.type== pygame.QUIT: #if the user presses the [x] in game window, it quits the window
            gameExit=True    

    if event.key == pygame.K_LEFT:
                lead_x_change = -block_size #block size is number of pixels moved in one loop
                lead_y_change=0

            elif event.key==pygame.K_RIGHT:
                lead_x_change=  block_size
                lead_y_change=0

            elif event.key==pygame.K_UP:
                lead_y_change= -block_size
                lead_x_change=0

            elif event.key==pygame.K_DOWN:
                lead_y_change= block_size
                lead_x_change=0
    lead_y+=lead_y_change
    lead_x+=lead_x_change

   gameDisplay.fill(white) #fills the display surface object, backgroud color is the parameter filled in        

   pygame.draw.rect(gameDisplay,black,[lead_x,lead_y,block_size,block_size])

   pygame.display.update() #after done with all the action,update the surface

   clock.tick(FPS) #runs the game at 30FPS

pygame.quit()
quit()

Pygame，在屏幕上移动一个方块，但不能擦除之前的移动

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