我在聚焦另一个进程id的窗口时遇到问题。
当它最小化时,我有问题,我通过测试IsIconic然后执行ShowWindow修复了这个问题。但是现在,如果进程不是前台进程,则不会显示。
SetForeground文档中写道:
关于如何将进程带到前台,有什么想法吗?
我看到了这个方法:
EnumWindows((WNDENUMPROC)topenumfunc, processid) ;此处使用:http://www.mombu.com/microsoft/windows-programmer-win32/t-bring-process-to-foreground-page2-1656417.html
但是,这会枚举所有窗口并回调测试当前窗口pid是否等于提供给EnumWindows函数的processid的pid。我只想在一个窗口上运行。
这是我的FocusWindow函数:
Cu.import('resource://gre/modules/ctypes.jsm');
var user32 = ctypes.open('user32.dll');
/* http://msdn.microsoft.com/en-us/library/ms633539%28v=vs.85%29.aspx
* BOOL WINAPI SetForegroundWindow(
* __in HWND hWnd
* );
*/
var SetForegroundWindow = user32.declare('SetForegroundWindow', ctypes.winapi_abi, ctypes.bool,
ctypes.int32_t
);
/* http://msdn.microsoft.com/en-us/library/windows/desktop/ms633522%28v=vs.85%29.aspx
* DWORD WINAPI GetWindowThreadProcessId(
* __in_ HWND hWnd,
* __out_opt_ LPDWORD lpdwProcessId
* );
*/
var GetWindowThreadProcessId = user32.declare('GetWindowThreadProcessId', ctypes.winapi_abi, ctypes.unsigned_long, //DWORD
ctypes.int32_t, //HWND
ctypes.unsigned_long.ptr //LPDWORD
);
/* http://msdn.microsoft.com/en-us/library/windows/desktop/ms633507%28v=vs.85%29.aspx
* BOOL WINAPI IsIconic(
* __in HWND hWnd
* );
*/
var IsIconic = user32.declare('IsIconic', ctypes.winapi_abi, ctypes.bool, // BOOL
ctypes.int32_t // HWND
);
/* http://msdn.microsoft.com/en-us/library/windows/desktop/ms633507%28v=vs.85%29.aspx
* BOOL WINAPI ShowWindow(
* __in HWND hWnd
* __in INT nCmdShow
* );
*/
var ShowWindow = user32.declare('ShowWindow', ctypes.winapi_abi, ctypes.bool, // BOOL
ctypes.int32_t, // HWND
ctypes.int // INT
);
var SW_RESTORE = 9;
function FocusWindow(hwnd) {
if (IsIconic(hwnd)) {
console.warn('its minimized so un-minimize it');
//its minimized so unminimize it
var rez = ShowWindow(hwnd, SW_RESTORE);
if (!rez) {
throw new Error('Failed to un-minimize window');
}
}
var rez = SetForegroundWindow(hwnd);
if (!rez) {
console.log('could not set to foreground window for a reason other than minimized, maybe process is not foreground, lets try that now');
var cPid = ctypes.cast(ctypes.voidptr_t(0), ctypes.unsigned_long);
var rez = GetWindowThreadProcessId(hwnd, cPid.address());
if (!rez) {
throw new Error('Failed to get PID');
} else {
console.log('cPid=', cPid, uneval(cPid), cPid.toString());
console.log('trying to set pid to foreground process');
return false;
}
} else {
return rez;
}
}发布于 2014-08-10 20:43:58
假设有一个父进程和几个子进程,那么它的一个子进程必须以父进程id作为参数执行AllowSetForegroundWindow函数。
然后,父进程可以成功调用SetForegroundWindow。
https://stackoverflow.com/questions/25226480
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