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.
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();
}
}
You can also try!
string allUsers=Environment.GetEnvironmentVariable("ALLUSERSPROFILE")+ "\\Start Menu\\Programs";
In .NET 4 CommonStartMenu
was added to the Environment.SpecialFolder
enum, so you can use:
Environment.GetFolderPath(Environment.SpecialFolder.CommonStartMenu)
"All Users" resides in the ALLUSERSPROFILE environment variable:
C:\>dir "%ALLUSERSPROFILE%\Start Menu"
Volume in drive C is awesome
Volume Serial Number is 8C57-DB1A
Directory of C:\Documents and Settings\All Users\Start Menu
12/28/2009 10:27 PM <DIR> .
12/28/2009 10:27 PM <DIR> ..
12/28/2009 10:01 PM 1,566 Microsoft Update.lnk
02/23/2010 09:57 PM <DIR> Programs
12/28/2009 10:27 PM 1,563 Set Program Access and Defaults.lnk
12/28/2009 08:51 PM 398 Windows Catalog.lnk
12/28/2009 08:51 PM 1,507 Windows Update.lnk
4 File(s) 5,034 bytes
3 Dir(s) 64,214,460,416 bytes free
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