为什么我不能在Windsor中为多个接口注册一个类?

Question

我正在尝试注册所有与Windsor实现IProcess<T1, T2>接口的类。为了完成这个任务，我在安装程序中添加了以下代码:

        // Register all implemented process interfaces
        var procTypes = AppDomain.CurrentDomain
                                 .GetAssemblies()
                                 .SelectMany(x => x.GetTypes())
                                 .Where(x => x.IsDerivedFromOpenGenericType(typeof(IProcess<,>)))
                                 .ToList();
        foreach (var procType in procTypes)
            foreach (var procInterface in procType.GetInterfaces().Where(x => x.IsDerivedFromOpenGenericType(typeof(IProcess<,>))))
                container.Register(Component.For(procInterface).ImplementedBy(procType).LifeStyle.Transient);

我想注册的一个类如下:

public class PositionProcesses 
    : IProcess<CreatePositionParams, PositionDisplayViewModel>,
      IProcess<EditPositionParams, PositionDisplayViewModel>
{
}

第一个接口被正确注册，但是在注册由这个类实现的第二个接口时，我得到了以下错误:

Test method MyJobLeads.Tests.Controllers.PositionControllerTests.Windsor_Can_Resolve_PositionController_Dependencies threw exception: 
Castle.MicroKernel.ComponentRegistrationException: There is a component already registered for the given key MyJobLeads.DomainModel.Processes.Positions.PositionProcesses

在第一次循环迭代时，我的变量是:

+       procInterface   {Name = "IProcess`2" FullName = "MyJobLeads.DomainModel.Data.IProcess`2[[MyJobLeads.DomainModel.ProcessParams.Positions.CreatePositionParams, MyJobLeads.DomainModel, Version=1.0.0.0, Culture=neutral, PublicKeyToken=null],[MyJobLeads.DomainModel.ViewModels.Positions.PositionDisplayViewModel, MyJobLeads.DomainModel, Version=1.0.0.0, Culture=neutral, PublicKeyToken=null]]"}   System.Type {System.RuntimeType}
+       procType    {Name = "PositionProcesses" FullName = "MyJobLeads.DomainModel.Processes.Positions.PositionProcesses"}  System.Type {System.RuntimeType}

第二行:

+       procInterface   {Name = "IProcess`2" FullName = "MyJobLeads.DomainModel.Data.IProcess`2[[MyJobLeads.DomainModel.ProcessParams.Positions.EditPositionParams, MyJobLeads.DomainModel, Version=1.0.0.0, Culture=neutral, PublicKeyToken=null],[MyJobLeads.DomainModel.ViewModels.Positions.PositionDisplayViewModel, MyJobLeads.DomainModel, Version=1.0.0.0, Culture=neutral, PublicKeyToken=null]]"} System.Type {System.RuntimeType}
+       procType    {Name = "PositionProcesses" FullName = "MyJobLeads.DomainModel.Processes.Positions.PositionProcesses"}  System.Type {System.RuntimeType}

(这两个都来自VS调试器。

任何想法?

Answer 1

你应该使用基于约定的组件注册

BasedOnDescriptor processes = AllTypes.FromAssembly(assemblyWithProcesses)
    .BasedOn(typeof (IProcess<,>))
    .WithService.AllInterfaces()
    .Configure(x => x.LifeStyle.Transient);
container.Register(processes)

EDIT删除了@Krzysztof-kozmic提到的第一个样本

Answer 2

如果你有一个为多个服务注册的组件，我认为你必须手动命名每个注册。

见:http://docs.castleproject.org/Windsor.Registering-components-by-conventions.ashx Configuring_registration_13

container.Register(
    Component.For(procInterface)
             .ImplementedBy(procType)
             .LifeStyle.Transient
             .Named(component.Implementation.FullName
                 + "-"
                 + procInterface.Name)
    );

这应该按照类型的全名加上要为其注册的接口注册每个组件。