问类推Python中的is_pressed？EN我正在制作一个简单的LTK游戏，在通用Lisp，并希望让玩家的角色跳上按下空间。但是，该窗口正在不断更新，因此我不能等待用户输入。我想检查每个帧是否按下空格键。尽管谷歌搜索了很多，但我无法在Lisp中找到这样的方法。我发现的只是阻止程序流的传统I/O提示。一、数组方法 创建数组
while (1): if keyboard.is_pressed('1'): x1, y1 = pyautogui.position() # 返回鼠标的坐标 pos1Str = "Position1:" + str(x1).rjust(4) + ',' + str(y1).rjust(4) print(pos1Str) # 打印坐标 print(get_color(x1, y1)[0]) if keyboard.is_pressed('2'): x2, y2 = pyautogui...
下面是完整的代码实现: importkeyboardwhileTrue:ifkeyboard.is_pressed('q'):breakprint("程序已退出") 1. 2. 3. 4. 5. 6. 7. 总结 本文介绍了如何使用Python实现按键退出功能。通过导入keyboard模块并使用is_pressed()函数来检测按键事件,我们可以轻松地实现按键退出的功能。希望本文对刚入行的小白有所帮助,...
在Python中,可以通过break语句来跳出当前循环,从而终止程序的执行。 whileTrue:ifkeyboard.is_pressed(" "):breakelse:# 其他按键的处理# 退出程序 1. 2. 3. 4. 5. 6. 7. 三、实例演示 下面是一个完整的示例代码,实现了按空格键退出程序的功能。 importkeyboardwhileTrue:ifkeyboard.is_pressed(" "):brea...
Check if a Button is pressed: 检测按钮(Button)是否被按下: 0 按钮按下与释放 from gpiozero import Button button = Button(2) while True: if button.is_pressed: print("Button is pressed") else: print("Button is not pressed") Wait for a button to be pressed before continuing: 等待按钮被...
有的人是因为原先从事的行业薪水比软件行业低，然后跳到这个行业；也有的是觉得软件测试门槛低，毕业后做...
ifkey_pressed[pygame.K_a] or key_pressed[pygame.K_LEFT]: hero.move(cfg.SCREENSIZE,'left') ifkey_pressed[pygame.K_d] or key_pressed[pygame.K_RIGHT]: hero.move(cfg.SCREENSIZE,'right') # --随机生成食物 generate_food_count += 1 ...
ifzombie.isAlive==False: continue self.energy-=1 ifself.energy <=0: forzombieinself.zombies: zombie.isMeetnust=False self.kill() self.image=self.images[args[0]%len(self.images)] Sun类: 1 2 3 4 5 6 7 8 9 10 11 12 13
The Python extension is working on improving the behavior of sending code to the Python REPL (Shift+Enter) when no code has been explicitly selected to run. Previously, when you placed your cursor on a line of Python code and pressed Shift+Enter, the Python extension would send the exact ...
##while True:# if keyboard.is_pressed('space'):# print('space was pressed!')## This will use 100% of your CPU and print the message many times.# Do this insteadwhileTrue: keyboard.wait('space')print('space was pressed! Waiting on it again...')# or thiskeyboard.add_hotkey('...