我想等到按下'q'按钮
这是我的代码:
import keyboard
while True:
if keyboard.is_pressed('q'):
print('q is pressed')
break #finishing the loop
这是我得到的例外:
Exception in thread Thread-1:
Traceback (most recent call last):
File "C:\Python27\lib\threading.py", line 810, in __bootstrap_inner
self.run()
File "C:\Python27\lib\threading.py", line 763, in run
self.__target(*self.__args, **self.__kwargs)
File "C:\Python27\lib\site-packages\keyboard\__init__.py", line 134, in listen
_os_keyboard.listen(self.queue, _key_table.is_allowed)
File "C:\Python27\lib\site-packages\keyboard\_winkeyboard.py", line 423, in listen
keyboard_hook = SetWindowsHookEx(WH_KEYBOARD_LL, keyboard_callback, NULL, NULL)
ArgumentError: argument 2: <type 'exceptions.TypeError'>: expected CFunctionType instance instead of CFunctionType
问题是什么?