如何使用数组制作一个矩形，以便我可以同时在屏幕上显示多个？

Z时代
2024-01-10
分类：问答

标题很自我解释。我试图找出如何使用Python的这个俄罗斯方块游戏的数组做一个矩形。如何使用数组制作一个矩形，以便我可以同时在屏幕上显示多个？

下面的代码：

screen = pygame.display.set_mode((400,800)) 
#Rectangle Variables 
x = 200 
y = 0 
width = 50 
height = 50 
thickness = 5 
speed = 1 
#Colors 
red = (255,0,0) 
white = (255,255, 255) 
green = (0,255,0) 
blue = (0,0,255) 
while(True): 
    for event in pygame.event.get(): 
     if event.type == pygame.QUIT: 
      pygame.quit(); sys.exit(); 
    #These lines^make the user able to exit out of the game window 
    y = y+1 
    pygame.draw.rect((screen) , red, (x,y,width,height), thickness) 
    pygame.display.update()

回答：

如果你想要做的就是添加矩形到一个数组，你可能只是这样做：

rectangles = [] 
while(True): 
    for event in pygame.event.get(): 
     if event.type == pygame.QUIT: 
      pygame.quit(); sys.exit(); 
    #These lines^make the user able to exit out of the game window 
    y = y+1 
    rectangles.append(pygame.draw.rect((screen) , red, (x,y,width,height), thickness)) 
    pygame.display.update()

回答：

如果你有一个位置列表，然后使用for环路画出来。

这里的位置是在像素

# --- constants --- (UPPER_CASE_NAMES) 
WIDTH = 50 
HEIGHT = 50 
RED = (255,0,0) 
# --- main --- 
rectangles_XY = [ (0, 0), (50, 0), (100, 0) ] 
while True: 
    for event in pygame.event.get(): 
     if event.type == pygame.QUIT: 
      # PLEASE, don't put all in one line 
      # it makes code less readable. 
      pygame.quit() 
      sys.exit() 
    for x, y in rectangles_XY: 
     pygame.draw.rect(screen, RED, (x, y, WIDTH, HEIGHT), 0) 
    pygame.display.update()

这里位置在小区现在的位置（列，行）

# --- constants --- (UPPER_CASE_NAMES) 
WIDTH = 50 
HEIGHT = 50 
RED = (255,0,0) 
# --- main --- 
rectangles = [ (0, 0), (1, 0), (2, 0) ] 
while True: 
    for event in pygame.event.get(): 
     if event.type == pygame.QUIT: 
      # PLEASE, don't put all in one line 
      # it makes code less readable. 
      pygame.quit() 
      sys.exit() 
    for column, row in rectangles: 
     x = column * WIDTH 
     y = row * HEIGHT 
     pygame.draw.rect(screen, RED, (x, y, WIDTH, HEIGHT), 0) 
    pygame.display.update()