Windsor MixIn is a Singleton?

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庸人自扰
庸人自扰 2021-01-26 06:25

I have a MixIn that requires some state to operate.

I am registering it as so..

    container.Register(Component.For(Of ICat) _
                        .         


        
3条回答
  •  轻奢々
    轻奢々 (楼主)
    2021-01-26 07:07

    I think I resolved this.

    Instead of using Proxy.Mixins, I created a custom Activator()

    Public Class MixInActivator(Of T)
       Inherits Castle.MicroKernel.ComponentActivator.DefaultComponentActivator
    
      Public Sub New(ByVal model As Castle.Core.ComponentModel, ByVal kernel As Castle.MicroKernel.IKernel, ByVal OnCreation As Castle.MicroKernel.ComponentInstanceDelegate, ByVal OnDestruction As Castle.MicroKernel.ComponentInstanceDelegate)
        MyBase.New(model, kernel, OnCreation, OnDestruction)
      End Sub
    
      Protected Overrides Function InternalCreate(ByVal context As Castle.MicroKernel.CreationContext) As Object
    
        Dim obj As Object = MyBase.InternalCreate(context)
        If GetType(T).IsAssignableFrom(obj.GetType) = False Then
            Dim options As New Castle.DynamicProxy.ProxyGenerationOptions
            Dim gen As New Castle.DynamicProxy.ProxyGenerator
            options.AddMixinInstance(Kernel.Resolve(Of T))
            obj = gen.CreateInterfaceProxyWithTarget(Model.Service, obj, options)
        End If
        Return obj
     End Function
    End Class
    

    So now, the component is registered like this

     container.Register(Component.For(Of ICat) _
                         .ImplementedBy(Of Cat) _
                         .LifeStyle.Is(Castle.Core.LifestyleType.Transient) _
                         .Activator(Of MixInActivator(Of IMixin)))
    

    And IMixin is registered as follows

    container.Register(Component.For(Of IMixin) _
                           .ImplementedBy(Of MyMixin) _
                           .LifeStyle.Is(Castle.Core.LifestyleType.Transient) _
                           .Named("MyMixin"))
    

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