[Python知识库]Python基础：第014课——弹跳的小球

观看视频
本次课，我们将实现小球在碰到窗口的边缘时发生反弹，先来看一下图。
参考：pygame中的Rect对象

示例1：让小球动起来

下面的代码，球没有移动的原因就是move()方法错误，move（x，y）方法：返回按给定偏移量移动后的新矩形。x和y参数可以是任何整数值（正数或负数）。而原来的Rect对象并没有移动，需要将图像绘制在新的Rect对象中，这样才能产生动画

import pygame, sys

pygame.init()

screen = pygame.display.set_mode((600,480))

pygame.display.set_caption("小小工坊")

ball = pygame.image.load('pygame/images/ball.gif')
ball_rect = ball.get_rect()

speed = 1
while True:

    for e in pygame.event.get():
        if e.type == pygame.QUIT:
            pygame.quit()
            sys.exit()
    
    ball_rect.move(speed, speed)
    screen.fill((0,0,255))
    screen.blit(ball,ball_rect)

    pygame.display.update()

因此，程序中ball_rect.move(speed, speed)这句应改成：ball_rect = ball_rect.move(speed, speed)

import pygame, sys

pygame.init()

screen = pygame.display.set_mode((600,480))

pygame.display.set_caption("小小工坊")

ball = pygame.image.load('pygame/images/ball.gif')
ball_rect = ball.get_rect()

speed = 1
while True:

    for e in pygame.event.get():
        if e.type == pygame.QUIT:
            pygame.quit()
            sys.exit()
    
    ball_rect = ball_rect.move(speed, speed)
    screen.fill((0,0,255))
    screen.blit(ball,ball_rect)

    pygame.display.update()

通过修改，球一闪而过，速度太快，这时，我们需要用到time模块，设定每秒钟游戏的刷新频率，因此，应该添加：

fclock = pygame.time.clock()
……
fclock.tick(60)

示例2：让小球反弹

当小球碰到边缘的情况，就是绘制小球所在的Rect对象碰到边缘，有四种情况：

• top < =0：竖直方向的速度变为负值
• bottom > screenHeight： 竖直方向的速度变为正值
• left <= 0 ：水平方向的速度变为正值
• right >= screenWidth：水平方向的速度变为负值

因此，程序需要改为：

import pygame, sys

pygame.init()

screen = pygame.display.set_mode((600,480))

pygame.display.set_caption("小小工坊")

ball = pygame.image.load('pygame/images/ball.gif')
ball_rect = ball.get_rect()

speed = 1

fclock = pygame.time.Clock()
while True:

    for e in pygame.event.get():
        if e.type == pygame.QUIT:
            pygame.quit()
            sys.exit()
    
    ball_rect = ball_rect.move(speed, speed)
    if ball_rect.right >= 600:
        speed = -1
        ball_rect.right = 600
    if ball_rect.left <= 0:
        speed = 1
        ball_rect.left = 0
        
    if ball_rect.bottom >= 480:
        speed = -1
        ball_rect.bottom = 480

    if ball_rect.top <= 0:
        speed = 1
        ball_rect.top = 0
    screen.fill((0,0,255))
    screen.blit(ball,ball_rect)

    pygame.display.update()
    fclock.tick(200)

但是此时，小球一直在一条对角线上来回运动，这是因为：不论在水平或竖直方向，只要超出屏幕的边界，speed的值就会改变，而speed是水平和竖直两个方向的速度。这时就要考虑用两个值分别表示水平和竖直方向上的速度，这时可以考虑用列表：speed = [1, 1]。这里能否用元组speed = (1, 1)呢？

import pygame, sys

pygame.init()

screen = pygame.display.set_mode((600,480))

pygame.display.set_caption("小小工坊")

ball = pygame.image.load('pygame/images/ball.gif')
ball_rect = ball.get_rect()

speed = [1,1]

fclock = pygame.time.Clock()
while True:

    for e in pygame.event.get():
        if e.type == pygame.QUIT:
            pygame.quit()
            sys.exit()
    
    ball_rect = ball_rect.move(speed[0], speed[1])
    if ball_rect.right >= 600:
        speed[0] = -1
        ball_rect.right = 600
    if ball_rect.left <= 0:
        speed[0] = 1
        ball_rect.left = 0

    if ball_rect.bottom >= 480:
        speed[1] = -1
        ball_rect.bottom = 480

    if ball_rect.top <= 0:
        speed[1] = 1
        ball_rect.top = 0
    screen.fill((0,0,255))
    screen.blit(ball,ball_rect)

    pygame.display.update()
    fclock.tick(200)