Python 3.5 HookManager SystemError：PyEval_EvalFrameEx

2019年8月3日 447次阅读

我是新来的,我希望不要犯任何错误！

我正在尝试使这个简单的代码工作,我在Python 3.4 32位测试它并且它工作但我需要在Python 3.5.0 64位使用它,但我得到这个错误,我不知道如何修复.

import pythoncom, pyHook

def OnKeyboardEvent(event):
    key=chr(event.Ascii)
    print(key)

hm = pyHook.HookManager()
hm.KeyDown = OnKeyboardEvent
hm.HookKeyboard()
pythoncom.PumpMessages()

我在屏幕上打印出按键,然后出现此错误：

During handling of the above exception, another exception occurred:
Traceback (most recent call last):
File "C:\Python 3.5\lib\site-packages\pyHook\HookManager.py", line 348, in KeyboardSwitch
event = KeyboardEvent(msg, vk_code, scan_code, ascii, flags, time, hwnd, win_name)
File "C:\Python 3.5\lib\site-packages\pyHook\HookManager.py", line 208, in __init__
HookEvent.__init__(self, msg, time, hwnd, window_name)
SystemError: PyEval_EvalFrameEx returned a result with an error set
TypeError: an integer is required (got type NoneType)

我真的不知道该怎么办！

最佳答案 您的函数需要返回一个整数值：

def OnKeyboardEvent(event):
    key=chr(event.Ascii)
    print(key)
    return 0
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