inversion-of-control – Castle-Windsor是否通过XML配置支持For

此功能尚未在XML解释器中实现.但是通过工具添加对它的支持并不困难(显然,当想要添加现有配置解析器中不存在的其他功能时,此技术也很有用).

首先,我们添加一个工具,它将检测何时为类型创建处理程序,同时将注册任何转发的服务,以便它们指向现有的处理程序：

public class HandlerForwardingFacility : AbstractFacility
{
  IConversionManager conversionManager;

  protected override void Init()
  {
    conversionManager = (IConversionManager)Kernel.GetSubSystem(SubSystemConstants.ConversionManagerKey);
    Kernel.HandlerRegistered += new HandlerDelegate(Kernel_HandlerRegistered);      
  }

  void Kernel_HandlerRegistered(IHandler handler,ref bool stateChanged)
  {
    if (handler is ForwardingHandler) return;

    var model = handler.ComponentModel;

    if (model.Configuration == null) return;

    var forward = model.Configuration.Children["forward"];
    if (forward == null) return;

    foreach (var service in forward.Children)
    {
      Type forwardedType = (Type)conversionManager.PerformConversion(service,typeof (Type));
      Kernel.RegisterHandlerForwarding(forwardedType,model.Name);
    }
  }
}

然后我们当然需要在代码中使用它,对于这个例子,我将有一个支持两个独立服务的变异鸭/狗组件 – IDuck和IDog：

public interface IDog
{
  void Bark();
}

public interface IDuck
{
  void Quack();
}

public class Mutant : IDog,IDuck
{
  public void Bark()
  {
    Console.WriteLine("Bark");
  }

  public void Quack()
  {
    Console.WriteLine("Quack");
  }
}

现在实际配置容器：

<castle>
  <facilities>
    <facility id="facility.handlerForwarding" type="Example.Facilities.HandlerForwardingFacility,Example" />
  </facilities>
  <components>
    <component id="mutant" service="Example.IDog,Example" type="Example.Mutant,Example">
      <forward>
        <service>Example.IDuck,Example</service>
      </forward>
    </component>
  </components>
</castle>

现在我们可以愉快地执行这样的测试：

WindsorContainer container = new WindsorContainer(new XmlInterpreter());

  var dog = container.Resolve<IDog>();
  var duck = container.Resolve<IDuck>();

  Assert.AreSame(dog,duck);

希望这可以帮助.