opygame：试图扑出一个球，但被卡住了

Question

import pygame

Red = 255, 0, 0
Black= 0,0,0
rectXpos = 2
rectypos = 2
speed = 2
screenedgex = 500

pygame.init()

window = pygame.display.set_mode(size=(500, 500))

clock = pygame.time.Clock()
running = True
while running:
    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            running = False

        pygame.display.update()

        window.fill(Black)
        square = pygame.draw.rect(window, Red, [rectXpos, rectypos, 50, 50],2)
        rectXpos += 2

        if rectXpos < 500:
            rectXpos -= 2



        clock.tick(60)
        print(rectXpos)`enter code here`

所以我在做什么错？我尝试制作if语句来停止球并反转它，但它使球保持在窗口的边缘

Answer 1

这是完整的代码，我分开了x和y跳动，因此您可以使用其中任何一个，也可以更新代码一点，再加上一些额外的格式。

# Imports
import pygame

# Vars
Red = 255, 0, 0
Black= 0,0,0
rectXpos = 2
rectYpos = 2
rect_width = 50
rect_height = 50
screen_width = 500
screen_height = 500
block_x_direction = 1
block_y_direction = 1

# Setup Code
pygame.init()
window = pygame.display.set_mode(size=(screen_width, screen_height))
clock = pygame.time.Clock()
running = True

# Game Loop
########################################################
while running:
    # Event Loop
    ########################################################
    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            running = False


    # Game Code - Update
    ########################################################
    # Game Code - Update - Rect X Bounce
    if rectXpos + (rect_width)>= screen_width:
        block_x_direction = block_x_direction * -1
    rectXpos += 2 * block_x_direction
    # Game Code - Update - Rect Y Bounce
    if rectYpos + (rect_height)>= screen_height:
        block_y_direction = block_y_direction * -1
    rectYpos += 2 * block_y_direction

    # - Tick Game    
    clock.tick(60)

    # Game Code - Render
    ########################################################
    window.fill(Black)
    square = pygame.draw.rect(window, Red, [rectXpos, rectYpos, rect_width, rect_height],2)
    pygame.display.update()

    # Game Code - Debug Code
    ########################################################
    print(clock.tick)

Answer 2

我假设您要在移动鼠标时来回移动矩形。您在这里做错了两件事：1.更正此：if rectXpos > 500:，因为当X达到500时必须减少X2.到达rectXpos 501时，应改变方向，直到到达rectXpos 0但您降低了头寸，直到头寸大于500，因此它会停留在499到501之间正确的代码：

import pygame

Red = 255, 0, 0
Black= 0,0,0
rectXpos = 2
rectypos = 2
speed = 2
screenedgex = 500

pygame.init()

window = pygame.display.set_mode(size=(500, 500))

clock = pygame.time.Clock()
running = True
k=1                                #here is k used to indicate direction
while running:
    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            running = False

        pygame.display.update()

        window.fill(Black)
        square = pygame.draw.rect(window, Red, [rectXpos, rectypos, 50, 50],2)
        rectXpos +=  2*k                #here is addition of 2 in given direction

        if (rectXpos > 500) or (rectXpos < 0): #here is condition to change direction
            k=-k



        clock.tick(60)
        print(rectXpos)

Answer 3

您应该在位置上添加speed，并且当触摸边框时，您应该将speed更改为-speed。

您也可以使用pygame.Rect()来保持位置和大小-它具有.left和.right（及其他）属性，这些属性可能非常有用。您可以使用Rect绘制pygame.draw.rect()（或检查与其他Rect的碰撞）

import pygame

# --- constants --- (UPPER_CASE_NAMES)

RED = (255, 0, 0)
BLACK = (0, 0, 0)

WIDTH = 500
HEIGHT = 500

# --- main ---

speed = 10

pygame.init()

window = pygame.display.set_mode((WIDTH, HEIGHT))

item = pygame.Rect(0, 0, 50, 50)

clock = pygame.time.Clock()
running = True
while running:

    # - events -

    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            running = False

    # - updates - (without draws)

    item.x += speed

    if item.right >= WIDTH:
        speed = -speed
    if item.left <= 0:
        speed = -speed

    # - draws - (without updates)

    window.fill(BLACK)
    pygame.draw.rect(window, RED, item, 2)
    pygame.display.update()

    clock.tick(60)

# - end -
pygame.quit()