Programmatically access All Users Start Menu

Question

Does anyone know how to programmatically access the "All Users" Startup Menu?

In XP, located here:

C:\Documents and Settings\All Users\Start Menu\Programs\Startup

And in Windows 7, located here:

C:\ProgramData\Microsoft\Windows\Start Menu\Programs\Startup

Specifically, I've got a Setup and Deployment project, and I'd like to put a shortcut to the application in the Startup menu for all users so that the application is start whenever anyone logs in.

EDIT: I'm pretty sure this is where Brian got his answer from.

Answer 1

There is no constant defined for the normal way of Environment.GetFolderPath for the all users start menu, but you can do it this way by using the Win32 API SHGetSpecialFolderPath:

class Program
{
    [DllImport("shell32.dll")]
    static extern bool SHGetSpecialFolderPath(IntPtr hwndOwner,
       [Out] StringBuilder lpszPath, int nFolder, bool fCreate);
    const int CSIDL_COMMON_STARTMENU = 0x16;  // All Users\Start Menu

    static void Main(string[] args)
    {
        StringBuilder path = new StringBuilder(260);
        SHGetSpecialFolderPath(IntPtr.Zero, path, CSIDL_COMMON_STARTMENU, false);
        string s = path.ToString();
    }
}