用python的pygame,做一个按钮

用python的pygame,做一个按钮

唔...摁钮？是指开始游戏这类的么？如果是的话：
import pygame
pygame.init() #初始化pygame
screen=pygame.display.set_mode([640,480])  #窗口大小：640*480
screen.fill([255,255,255])#用白色填充窗口
myimage=pygame.image.load(‘某个用来做摁钮的图片’) #把变量myimage赋给导入的图片
screen.blit(myimage,[100,100]) #在100,100的地方画出这个图片（100和100为左部和上部）
pygame.display.flip() 
while True:
    for event in pygame.event.get():#获得事件
        if event.type==pygame.MOUSEBUTTONDOWN and 100<=event.pos[0]<=图片宽 and 
         100<=event/pos[1]<=图片长: #判断鼠标位置以及是否摁了下去。
            #做需要做的事情，如开始游戏。
            pass够详细得了！

呵呵，有点长。
import pygame
from pygame.locals import *

class SimpleUI:

RED = (255,0,0)
BLUE = (0,0,255)
def __init__(self):
# Initialize PyGame
pygame.init()
pygame.display.set_caption('Paint')
self.screen = pygame.display.set_mode((640,480))
self.screen.fill((255,255,255))
self.button1 = Rect(20, 20, 100, 50)
self.button2 = Rect(20, 90, 100, 50)
self.colorArea = Rect(140, 20, 400, 400)
self.selectedColor = SimpleUI.BLUE

def run(self):
# Run the event loop
self.loop()
# Close the Pygame window
pygame.quit()

def loop(self):
exit = False
while not exit:
exit = self.handleEvents()
self.draw()
def handleEvents(self):
exit = False
for event in pygame.event.get():
if event.type == QUIT:
exit = True
elif event.type == KEYDOWN:
if event.key == K_ESCAPE:
exit = True
elif event.type == MOUSEBUTTONDOWN:
self.handleMouseDown(pygame.mouse.get_pos())
return exit
def handleMouseDown(self, (x, y)):
print x,y
if (self.button1.collidepoint(x, y)):
self.selectedColor = SimpleUI.BLUE
print "BLUE"
elif (self.button2.collidepoint(x, y)):
self.selectedColor = SimpleUI.RED
print "RED"
def draw(self):
pygame.draw.rect(
self.screen,
SimpleUI.BLUE,
self.button1,
)
pygame.draw.rect(
self.screen,
SimpleUI.RED,
self.button2,
)
pygame.draw.rect(
self.screen,
self.selectedColor,
self.colorArea,
)
pygame.display.update()

# Start the game
SimpleUI().run()

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