在PyGame中，我需要将图片的坐标设置为变量（x，y）

Question

码：

import pygame

pygame.init()

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

blockImg = pygame.image.load('Rectangle.png')

block_rect = blockImg.get.rect()

x = block_rect.x   
/or x = block_rect.left/

y = block_rect.y   
/or y = block_rect.top/

print(x, y)

问题

当我编写一些代码以稳定的速度在屏幕上移动图像并不断更新图像的x和y时，它将仅打印出“（0，0）”，就像图像在左上角一样的窗口，不动

我究竟做错了什么？

Answer 1

很难知道您在这里做错了什么，因为您没有发布blit代码。

如果我不得不猜测的话，您可能没有更新在流向屏幕时使用的x，y变量。 他们不会自动更新自己，您必须设置

x = block_rect.left 

每个帧。

但是，这里有一些最小的工作代码可以满足您的期望。

import pygame

BGCOLOR = (100,100,100)
SCREENWIDTH = 600
SCREENHEIGHT = 400

pygame.init()

display = pygame.display.set_mode((SCREENWIDTH, SCREENHEIGHT))
clock = pygame.time.Clock()

block_img = pygame.image.load('Rectangle.png')
block_rect = block_img.get_rect()

#set velocity variable for updating position of rect
#make sure you do this before you go into the loop
velocity = 1

while 1:
    #fill in display
    display.fill(BGCOLOR)

    #pygame event type pygame.QUIT is activated when you click the X in the topright 
    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            #if you don't call pygame.quit() sometimes python will crash
            pygame.quit()
            #exit loop
            break

    #reverses velocity variable if rect goes off screen
    if block_rect.right > SCREENWIDTH:
        velocity = -velocity
    elif block_rect.left < 0:
        velocity = -velocity

    #update blocks position based on velocity variable
    block_rect.left += velocity

    #variables representing the x, y position of the block
    x, y = block_rect.left, block_rect.top
    display.blit(block_img, (x,y))

    #display.blit(block_img, block_rect) would also work here
    pygame.display.flip()
    clock.tick(60)