Castle Windsor - 传递类型作为构造函数参数

Question

我有一个类需要Type作为构造函数参数。我如何在Windsor容器中注册？

public Class NLogLogger : ILogger 
    { 
     private NLog.Logger _logger; 
     public NLogLogger(Type type) 
     { 
      _logger = NLog.LogManager.GetLogger(type.FullName); 
     } 
    ... 
    } 

我想这种方式进行注册：

  container.Register(Component.For(typeof(ILogger)) 
           .ImplementedBy(typeof(NLogLogger)) 
           .LifestyleTransient() 
           .DependsOn(new Hashtable 
              { 
               {"type", ???} 
              })); 

我需要填写???。我无法弄清楚如何在那里传递呼叫班的Type。

Answer 1

更新

通过您的更新去要注入的类型，这样就可以用它来记录。有两种解决方案。

public class ClassDependentOnLogger 
{ 
    private ILogger _logger; 
    public ClassDependentOnLogger(ILogger logger) 
    { 
     _logger = logger; 
    } 

    .... 

} 

1 - 使用温莎SubDependancyResolver。下面的例子演示了如何恢复和ILog的实例，但你可以很容易地适应它返回一个类型，而不是如果这是你想做的事：

public class LoggerResolver : ISubDependencyResolver 
{ 
    private readonly IKernel kernel; 

    public LoggerResolver(IKernel kernel) 
    { 
     this.kernel = kernel; 
    } 

    public object Resolve(CreationContext context, ISubDependencyResolver contextHandlerResolver, Castle.Core.ComponentModel model, DependencyModel dependency) 
    { 
     return NLog.LogManager.GetLogger(model.Implementation.FullName); 
    } 

    public bool CanResolve(CreationContext context, ISubDependencyResolver contextHandlerResolver, Castle.Core.ComponentModel model, DependencyModel dependency) 
    { 
     return dependency.TargetType == typeof(ILogger); 
    } 
} 

//Register the sub dependency resolver. This looks cleaner if you do it via a 
//Facility but that's a whole other class 
Container.Kernel.Resolver.AddSubResolver(new LoggerResolver(Kernel)); 

2 - 使用温莎LoggingFacility

有不需要你的例子中描述的NLoggerClass。