Question

这似乎是一个非常容易回答的问题，但我只是一个需要快速帮助的初学者。

我正在尝试创建一个程序，当你点击pyGame窗口的某个地方时，它会打印出你用鼠标左键点击它，并打印出按下它的位置的坐标。我已经有了这个。我在绘制pyGame窗口上的像素时遇到问题。基本上，我希望它绘制一个像素，我按下了pyGame窗口。

#!/usr/bin/env python

#import the module for use 
import pygame

#setting up some variables
running = 1
LEFT = 1 
#Set up the graphics area/screen
screen=pygame.display.set_mode((640,400))

#continuous loop to keep the graphics running
while running==1:
    event=pygame.event.poll()
    if event.type==pygame.QUIT:
        running=0
        pygame.quit()
    elif event.type==pygame.MOUSEBUTTONDOWN and event.button==LEFT:
        print "You pressed the left mouse button at (%d,%d)" %event.pos
    elif event.type==pygame.MOUSEBUTTONUP and event.button==LEFT:
        print "You released the left mouse button at (%d,%d)" %event.pos

Answer 1

尝试在收到鼠标按下事件时设置每个像素的颜色。

elif event.type==pygame.MOUSEBUTTONDOWN and event.button==LEFT:
  print "You pressed the left mouse button at (%d,%d)" %event.pos
  screen.set_at((event.pos.x, event.pos.y), pygame.Color(255,0,0,255))

请注意，这会根据需要临时锁定和解锁Surface。

Answer 2

我将Esthete的代码编辑为一个独立的工作示例：

根据您的操作，获取/设置单个像素可能会很慢。（如果需要，可以使用Surfarray和Pixelarray。）

import pygame
from pygame.locals import *

class Game(object):
    done = False
    def __init__(self, width=640, height=480):
        pygame.init()
        self.width, self.height = width, height
        self.screen = pygame.display.set_mode((width, height))

        # start with empty screen, since we modify it every mouseclick
        self.screen.fill(Color("gray50"))

    def main_loop(self):
        while not self.done:
            # events
            events = pygame.event.get()
            for event in events:
                if event.type == pygame.QUIT: self.done = True
                elif event.type == KEYDOWN:
                    if event.key == K_ESCAPE: self.done = True
                elif event.type == MOUSEMOTION:
                    pass
                elif event.type == MOUSEBUTTONDOWN and event.button == 1:
                    print "Click: ({})".format(event.pos)
                    self.screen.set_at(event.pos, Color("white"))

            # draw
            pygame.display.flip()


if __name__ == "__main__":
    g = Game()
    g.main_loop()

PyGame - 绘制像素并打印出它的坐标

2 个答案: