Question

我想在python中使用win32api模块安装一个hook，并捕获wm_activate消息，以便对其进行进一步处理，然后将消息发布给其他应用程序进行处理。基本上我想检测用户是否更改了活动的前景窗口。我已经使用 PySide2 模块中的 QTimer 每 25 毫秒检查活动前景窗口是否已更改，但这不是最佳解决方案。

这是我的尝试，但我没有从中得到任何消息，因为从不执行 low_level_handler 回调函数。它适用于低级鼠标钩子。我错过了什么？

# Some of this adapted from BoppreH's answer here:http://stackoverflow.com/questions/9817531/applying-low-level-keyboard-hooks-with-python-and-setwindowshookexa
import win32con, win32gui
from ctypes import wintypes
from ctypes import windll, CFUNCTYPE, POINTER, c_int, c_void_p, byref, WINFUNCTYPE 
import atexit

Handlers=[]

def listener(yo):
    def low_level_handler(nCode, wParam, lParam):
        print("reach")
        print(nCode,wParam, lParam)
        #Be nice, return next hook
        return windll.user32.CallNextHookEx(hook_id, nCode, wParam, lParam)
    
    # Our low level handler signature.
    CMPFUNC = WINFUNCTYPE(c_int, c_int, c_int, POINTER(c_void_p))
    # Convert the Python handler into C pointer.
    pointer = CMPFUNC(low_level_handler)

    windll.kernel32.GetModuleHandleW.restype = wintypes.HMODULE
    windll.kernel32.GetModuleHandleW.argtypes = [wintypes.LPCWSTR]
    windll.user32.SetWindowsHookExA.argtypes = (c_int, wintypes.HANDLE, wintypes.HMODULE, wintypes.DWORD)
    print(yo)
    hwnd = windll.kernel32.GetModuleHandleW(None)
    hook_id = windll.user32.SetWindowsHookExA(win32con.WH_CALLWNDPROC, pointer, hwnd, 0)
    HOOK_ID = hook_id


    # Register to remove the hook when the interpreter exits.
    atexit.register(windll.user32.UnhookWindowsHookEx, hook_id)
    while True:
        windll.user32.PostThreadMessageW(hook_id, win32con.WM_ACTIVATE, 0, 0)
        msg = win32gui.GetMessage(None, 0, 0)
        win32gui.TranslateMessage(byref(msg))
        win32gui.DispatchMessage(byref(msg))

WM_QUIT = 0x0012
HOOK_ID = None

def installHook():
    listener("yo")

def removeHook():
    windll.user32.UnhookWindowsHookEx(HOOK_ID)
    HOOK_ID = None

installHook()

如何在python中捕获wm_activate消息

0 个答案: