2016-04-28 41 views
0

我为IndoorAtlas创建了一个绑定库。除了加载平面图之外,我迄今在创建应用程序方面都很成功。我用抽象方法fetchFloorPlanWithId专门解决了生成的C#类的问题。 SDK文档描述如下:abstract IATask<IAFloorPlan> fetchFloorPlanWithId(String floorPlanId)在Xamarin绑定库中删除类型

但是,生成的C#方法是 public abstract IATask FetchFloorPlanWithId(string p0);

我想铸造它作为 (IATask<IAFloorPlan>) mResourceManager.FetchFloorPlanWithId(id)但我得到一个无效的转换错误

我需要帮助解决这个问题指向我正确的方向。

下面是用于IAResourceManager

[Register("com/indooratlas/android/sdk/resources/IAResourceManager", DoNotGenerateAcw = true)] 
public abstract class IAResourceManager : Java.Lang.Object 
{ 
    [Register(".ctor", "()V", "")] 
    public IAResourceManager(); 
    protected IAResourceManager(IntPtr javaReference, JniHandleOwnership transfer); 

    protected override IntPtr ThresholdClass { get; } 
    protected override Type ThresholdType { get; } 

    [Register("create", "(Landroid/content/Context;)Lcom/indooratlas/android/sdk/resources/IAResourceManager;", "")] 
    public static IAResourceManager Create(Context p0); 
    [Register("create", "(Landroid/content/Context;Landroid/os/Bundle;)Lcom/indooratlas/android/sdk/resources/IAResourceManager;", "")] 
    public static IAResourceManager Create(Context p0, Bundle p1); 
    [Register("fetchFloorPlanWithId", "(Ljava/lang/String;)Lcom/indooratlas/android/sdk/resources/IATask;", "GetFetchFloorPlanWithId_Ljava_lang_String_Handler")] 
    public abstract IATask FetchFloorPlanWithId(string p0); 
} 

回答

0

解决所产生的C#类。我通过直接编辑绑定库项目中生成的C#类和接口来解决此问题