Die korrekte Methode zum Starten einer AppX-App vom Desktop besteht darin, die Methode IApplicationActivationManager::ActivateApplication
der Shell zu verwenden.
void Main()
{
var mgr = new ApplicationActivationManager();
uint processId;
var name = "Microsoft.WindowsSoundRecorder_8wekyb3d8bbwe!App";
mgr.ActivateApplication(name, null, ActivateOptions.None, out processId);
}
public enum ActivateOptions
{
None = 0x00000000, // No flags set
DesignMode = 0x00000001, // The application is being activated for design mode, and thus will not be able to
// to create an immersive window. Window creation must be done by design tools which
// load the necessary components by communicating with a designer-specified service on
// the site chain established on the activation manager. The splash screen normally
// shown when an application is activated will also not appear. Most activations
// will not use this flag.
NoErrorUI = 0x00000002, // Do not show an error dialog if the app fails to activate.
NoSplashScreen = 0x00000004, // Do not show the splash screen when activating the app.
}
[ComImport, Guid("2e941141-7f97-4756-ba1d-9decde894a3d"), InterfaceType(ComInterfaceType.InterfaceIsIUnknown)]
interface IApplicationActivationManager
{
// Activates the specified immersive application for the "Launch" contract, passing the provided arguments
// string into the application. Callers can obtain the process Id of the application instance fulfilling this contract.
IntPtr ActivateApplication([In] String appUserModelId, [In] String arguments, [In] ActivateOptions options, [Out] out UInt32 processId);
IntPtr ActivateForFile([In] String appUserModelId, [In] [MarshalAs(UnmanagedType.Interface, IidParameterIndex = 2)] /*IShellItemArray* */ IShellItemArray itemArray, [In] String verb, [Out] out UInt32 processId);
IntPtr ActivateForProtocol([In] String appUserModelId, [In] IntPtr /* IShellItemArray* */itemArray, [Out] out UInt32 processId);
}
[ComImport, Guid("45BA127D-10A8-46EA-8AB7-56EA9078943C")]//Application Activation Manager
class ApplicationActivationManager : IApplicationActivationManager
{
[MethodImpl(MethodImplOptions.InternalCall, MethodCodeType = MethodCodeType.Runtime)/*, PreserveSig*/]
public extern IntPtr ActivateApplication([In] String appUserModelId, [In] String arguments, [In] ActivateOptions options, [Out] out UInt32 processId);
[MethodImpl(MethodImplOptions.InternalCall, MethodCodeType = MethodCodeType.Runtime)]
public extern IntPtr ActivateForFile([In] String appUserModelId, [In] [MarshalAs(UnmanagedType.Interface, IidParameterIndex = 2)] /*IShellItemArray* */ IShellItemArray itemArray, [In] String verb, [Out] out UInt32 processId);
[MethodImpl(MethodImplOptions.InternalCall, MethodCodeType = MethodCodeType.Runtime)]
public extern IntPtr ActivateForProtocol([In] String appUserModelId, [In] IntPtr /* IShellItemArray* */itemArray, [Out] out UInt32 processId);
}
[ComImport]
[InterfaceType(ComInterfaceType.InterfaceIsIUnknown)]
[Guid("43826d1e-e718-42ee-bc55-a1e261c37bfe")]
interface IShellItem
{
}
[ComImport]
[InterfaceType(ComInterfaceType.InterfaceIsIUnknown)]
[Guid("b63ea76d-1f85-456f-a19c-48159efa858b")]
interface IShellItemArray
{
}
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Dank, versuchen und Feedback zu teilen. Schnelle Frage - Ich möchte, dass die Windows Form-Anwendung von einem normalen Benutzer ausgeführt wird. Ich hoffe, dass Ihr Code keine (erhöhten) Admin-Berechtigungen erfordert. – kamleshrao
Es aktiviert die App-Instanz für den aktuellen Benutzer. Keine Admin-Eingabeaufforderung. – codekaizen
@kamlesrao Hat das für dich funktioniert? Wenn ja, bitte upvote und akzeptiere als Antwort. – gravity