Question

我正在开发一个程序，它将显示一个数组中的字母。如果用户按下向上箭头，这将在阵列中移动，如果用户按下向下箭头。问题是，当我按下这些键中的任何一个时，字母移动得非常快，例如，如果我在A中，程序会移动到D或类似的东西。这是我的代码

def HighScore(event):
    pygame.init()
    nameString=""
    nameStringFinal=""
    index=0
    ABC=["A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"]
    font=pygame.font.SysFont("Arial",30,True,False)
    screen=pygame.display.set_mode((700, 700))
    pygame.display.set_caption("New High Score" )
    font=pygame.image.load("imagenes/paisaje.jpg").convert()
    escape=False        
    while escape==False:
        for event in pygame.event.get():
            if event.type == pygame.QUIT:
                pygame.quit()
                sys.exit()
        screen.blit(fondo,(0,0))
        show=fuente.render(nombreString,0,(0,0,0))

        if event.type ==  pygame.KEYDOWN:
            if control==True:
                if event.key == pygame.K_UP:  
                    if index==25:
                        index=25
                    else:
                        index=index+1
                    nameString=ABC[index]
                elif event.key == pygame.K_DOWN:
                    if index==0:
                        index=0
                    else:
                        index=index-1
                    nameString=ABC[index]
                elif event.key == pygame.K_RETURN:
                    nameStringFinal=nameStringFinal+nameString
        screen.blit(despliegue,(200,200))
        pygame.display.flip()

Answer 1

您需要确保所有事件处理都正确缩进，因此它是for循环的一部分

现在你的方式，最后的偶数将被反复处理，直到新的事件到来

例如：

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

    if event.type ==  pygame.KEYDOWN:
        if event.key == pygame.K_UP:  
            if index==25:
                index=25
            else:
                index=index+1

            nameString=ABC[index]
        elif event.key == pygame.K_DOWN:
            if index==0:
                index=0
            else:
                index=index-1
            nameString=ABC[index]
        elif event.key == pygame.K_RETURN:
            nameStringFinal=nameStringFinal+nameString

screen.blit(fondo,(0,0))
show=fuente.render(nombreString,0,(0,0,0))
screen.blit(despliegue,(200,200))

在shell中，你可以看到for循环在空列表上的方式，只是将变量保留为前一个循环的值

>>> for i in [1,2,3]:   # imagine this is the event list
...     if i == 4:
...         sys,exit()
... 
>>> if i == 3:
...     print "i==3"
... 
i==3
>>> for i in []:        # next time round the while loop, the event list is empty
...     if i == 4:
...         sys,exit()
... 
>>> if i == 3:          # but you're still processing this event again!
...     print "i==3"
... 
i==3

在pygame中管理keydown

1 个答案: