Show StartMenu on custom position in C#

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Hi together.

I have a little problem. I want to show the startmenu of win7 on a custom position but my code did nothing.

C#

IntPtr StartMenu = FindWindow("DV2ControlHost", null);
MoveWindow(StartMenu, x, y, width, height, true);
ShowWindowAsync(StartMenu, 1);

Could anyone help me?

Posted 20-Jun-11 4:54am

PotensGladius

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Sergey Alexandrovich Kryukov · Answer 1 · 2011-06-20T05:06:00

Why do you think it should show the Start Menu?

You can easily simulate Ctrl+Esc (which normally shows this menu) by using P/Invoked Windows API SendInput, see http://msdn.microsoft.com/en-us/library/ms646310(v=vs.85).aspx[^].

However I would not recommend doing anything about Start Menu at all. A good user will feel abused. The user know when and why using Start menu. What's the purpose of it? If you need to start something, this is done via System.Diagnostic.Process.Start, see http://msdn.microsoft.com/en-us/library/system.diagnostics.process.aspx[^].

—SA