2016-07-26 4 views

Antwort

1

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 
{ 
} 

Download this code for LinqPad

+0

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

+1

Es aktiviert die App-Instanz für den aktuellen Benutzer. Keine Admin-Eingabeaufforderung. – codekaizen

+1

@kamlesrao Hat das für dich funktioniert? Wenn ja, bitte upvote und akzeptiere als Antwort. – gravity

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