Python 2.7 windows钩子必须与消息泵注册在同一线程上吗？

我试图利用Windows钩子在应用程序发送自己的gui事件时拦截和阻止击键

我提出了以下清单：

import pythoncom
import pyHook
import threading
import time


def on_keyboard_event(event):
    print 'MessageName:',event.MessageName
    print 'Message:',event.Message
    print 'Time:',event.Time
    print 'Window:',event.Window
    print 'WindowName:',event.WindowName
    print 'Ascii:', event.Ascii, chr(event.Ascii)
    print 'Key:', event.Key
    print 'KeyID:', event.KeyID
    print 'ScanCode:', event.ScanCode
    print 'Extended:', event.Extended
    print 'Injected:', event.Injected
    print 'Alt', event.Alt
    print 'Transition', event.Transition
    print '---'
    return False


class WindowsHooksWrapper:
    def __init__(self):
        self.started = False
        self.thread = threading.Thread(target=self.thread_proc)
        self.hook_manager = pyHook.HookManager()

    def start(self):
        if self.started:
            self.stop()

        # Register hook
        self.hook_manager.KeyDown = on_keyboard_event
        self.hook_manager.KeyUp = on_keyboard_event
        self.hook_manager.HookKeyboard()

        # Start the windows message pump
        self.started = True
        self.thread.start()

    def stop(self):
        if not self.started:
            return

        self.started = False
        self.thread.join()

        self.hook_manager.UnhookKeyboard()

    def thread_proc(self):
        print "Thread started"
        while self.started:
            pythoncom.PumpWaitingMessages()

        print "Thread exiting..."


class WindowsHooksWrapper2:
    def __init__(self):
        self.started = False
        self.thread = threading.Thread(target=self.thread_proc)

    def start(self):
        if self.started:
            self.stop()

        self.started = True
        self.thread.start()

    def stop(self):
        if not self.started:
            return

        self.started = False
        self.thread.join()

    def thread_proc(self):
        print "Thread started"

        # Evidently, the hook must be registered on the same thread with the windows msg pump or
        #     it will not work and no indication of error is seen
        # Also note that for exception safety, when the hook manager goes out of scope, it
        #     unregisters all outstanding hooks
        hook_manager = pyHook.HookManager()
        hook_manager.KeyDown = on_keyboard_event
        hook_manager.KeyUp = on_keyboard_event
        hook_manager.HookKeyboard()

        while self.started:
            pythoncom.PumpWaitingMessages()

        print "Thread exiting..."
        self.hook_manager.UnhookKeyboard()


def main():
    # hook_manager = pyHook.HookManager()
    # hook_manager.KeyDown = on_keyboard_event
    # hook_manager.KeyUp = on_keyboard_event
    # hook_manager.HookKeyboard()
    # pythoncom.PumpMessages()

    hook_wrapper = WindowsHooksWrapper2()
    hook_wrapper.start()
    time.sleep(30)
    hook_wrapper.stop()


if __name__ == "__main__":
    main()

主要的注释部分来自pyhook wiki教程，效果很好

然后我尝试将其集成到一个类中，即“WindowsHooksRapper”类。如果我使用该类，它将不起作用，键盘消息将传递到它们的预期目标

凭直觉，我尝试了“WindowsHooksRapper2”，我将钩子的注册移动到消息泵的同一个线程。它现在起作用了

我的直觉正确吗？注册要求与泵在同一螺纹上？若然，原因为何

---

注意到，这是Windows 32 API的要求，而不是Python或PyHoover库本身，因为我在C++中使用了，并且使用“StWistWORSRead”直接得到了相同的结果。

< p>您创建了一个线程范围钩子。 这些钩子事件要么与特定线程相关联，要么与调用线程所在桌面上的所有线程相关联

Python中的

pythoncom.PumpWaitingMessages（）

和Win32中的

GetMessage

/

PeekMessage

是从“特定线程或与调用线程位于同一桌面上的所有线程”获取消息的方法

要创建全局钩子，为了让所有进程都可以访问键盘钩子，必须将其放置在DLL中，然后将其加载到每个进程的地址空间中。有关如何制作全局键盘挂钩的详细信息，请参见此