我需要帮助如何摆脱这种情况？

如何解决我需要帮助如何摆脱这种情况？

每当我运行我的代码时，我都会收到此错误：

 File "C:\Users\HOC\Desktop\Trial files\T1.py",line 50,in <module>
    screen.blit(font.render(text,True,(0,0)),(300,180))
AttributeError: 'tuple' object has no attribute 'blit'

我不知道如何处理这个问题。这是我正在制作的游戏，您可以在其中回答问题以在比赛中进步。我打算添加按钮和选项供玩家选择，但我什至无法通过我的按钮进行倒计时。

    import.pygame

    pygame.init()
    screen = pygame.display.set_mode((1000,1000))
    screen = screen_width,screen_height = 1000,1000

    def button(screen,position,text):
    font = pygame.font.SysFont("Times New Romans",50)
    text_render = font.render(text,1,(255,0))
    x,y,w,h = text_render.get_rect()
    x,y = position
    pygame.draw.line(screen,(150,150,150),(x,y),(x + w,5)
    pygame.draw.line(screen,y - 2),y + h),(50,50,50),y+h),[x + w,y],5)
    pygame.draw.rect(screen,(100,100,100),h))
    return screen.blit(text_render,y))

    def start():
    print("Ok,let's go")

    def menu():
        b1 = button(screen,(400,300),"Quit")
        b2 = button(screen,(500,"Start")
    counter,text = 3,'3'.rjust(3)
    pygame.time.set_timer(pygame.USEREVENT,1000)
    font = pygame.font.SysFont('Times New Romans',30)
    run = True
    while run:
        for count in pygame.event.get():
            if count.type == pygame.USEREVENT:
                counter -= 1
            text = str(counter).rjust(3) if counter > 0 else 'GO GO GO!'
        if count.type == pygame.QUIT: 
            run = False
        for event in pygame.event.get():
            if (event.type == pygame.QUIT):
                pygame.quit()
            if event.type == pygame.KEYDOWN:
                if event.key == pygame.K_ESCAPE:
                    pygame.quit()
            if event.type == pygame.MOUSEBUTTONDOWN:
                if b1.collidepoint(pygame.mouse.get_pos()):
                    pygame.quit()
                elif b2.collidepoint(pygame.mouse.get_pos()):
                    start()
        pygame.display.update()
        pygame.quit()
        screen.blit(font.render(text,180))
        pygame.display.flip()
        clock.tick(60)
    

menu()

解决方法

问题出在这里：

{"brand" : "A","productType" : "food","infoList" : [
            "pizza-34","cake-qaw-34"
          ]}

您正在用元组 (1000,1000) 覆盖屏幕对象。将此行更改为：

screen = screen_width,screen_height = 1000,1000

可能会解决您的问题。