在最顶层创建一个流程窗口 [英] create a process window on topmost

问题描述

在我的主应用程序中，我调用一个名为A的进程。这个进程正在调用另一个应用程序B.问题是，进程A始终位于我的主窗口顶部，但是当进程B启动时它位于我的主要后面window.It不是每次我都在进行操作，但有时候。



调用应用程序A后，我已申请等待退出然后我正在制作窗口线程要睡觉（这在我的情况下是必需的），同时应用程序B由A.启动。这个B进程窗口有时在主窗口后面。



任何人都可以建议我做任何事情吗？

In my main application,I'mm calling a process named A.This process is calling another application B.Problem is,the process A is always top on my main window,but when process B starts it is situated behind my main window.It is occurring not every time I am doing the operation,but sometime.

After calling application A,I have applied wait for exit and then I am making main window thread to sleep(this is required in my case),in the mean time application B is started by A.This B process window sometime is behind the main window.

can any one suggest me to do anything?

推荐答案

请注意，这是MSDN的解决方案的副本，您的特定情况可能不需要cx和cy ...



如果windows表格请试试这个：



处理流程= Process.Start（psi） ;

IntPtr windowHandle = process.MainWindowHandle;



RECT r;

GetWindowR ect（new HandleRef（this，windowHandle），out r）;



int cx = r.Right - r.Left;

int cy = r.Bottom - r.Top;



cx - = 20;

cy - = 20;



SetWindowPos（windowHandle，（IntPtr）SpecialWindowHandles.HWND_TOPMOST，r.Left，r.Top，cx，cy，SetWindowPosFlags.SWP_SHOWWINDOW）;





如果这有帮助，请花时间接受解决方案。

Note that this is a copy of a solution from MSDN, cx and cy may not be needed for your particular case...

Try this if windows forms:

Process process = Process.Start(psi);
IntPtr windowHandle = process.MainWindowHandle;

RECT r;
GetWindowRect(new HandleRef(this, windowHandle), out r);

int cx = r.Right - r.Left;
int cy = r.Bottom - r.Top;

cx -= 20;
cy -= 20;

SetWindowPos(windowHandle, (IntPtr) SpecialWindowHandles.HWND_TOPMOST, r.Left, r.Top, cx, cy, SetWindowPosFlags.SWP_SHOWWINDOW);


If this helps, please take time to accept the solution.

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