Question

如何在SimpleInjector中注册条件装饰器？以下是我的定义：

public interface ICommand { }

public interface ICleanableCommand : ICommand {
    void Clean();
}

public interface ICommandHandler<in TCommand> 
    where TCommand : ICommand {
    void Handle(TCommand command);
}

public class CleanableCommandHandlerDecorator<TCommand> 
    : ICommandHandler<TCommand> 
    where TCommand : ICleanableCommand {

    private readonly ICommandHandler<TCommand> _handler;

    public CleanableCommandHandlerDecorator(
        ICommandHandler<TCommand> handler) {
        _handler = handler;
    }

    void ICommandHandler<TCommand>.Handle(TCommand command) {
        command.Clean();
        _handler.Handle(command);
    }
}

我正在努力：

container.RegisterManyForOpenGeneric(
    typeof(ICommandHandler<>),
    AppDomain.CurrentDomain.GetAssemblies()
    );

container.RegisterDecorator(
    typeof(ICommandHandler<>),
    typeof(CleanableCommandHandlerDecorator<>)
    // ,context => context.ImplementationType ???
    // I want to register this decorator for commandhandlers 
    // which their command implements ICleanableCommand 
    );

Answer 1

您可以使用RegisterDecorator重载来DecoratorPredicateContext来定义应用装饰器的条件。但是，因为在您的情况下，条件只是泛型类型约束，所以您不必提供谓词。当给定的服务类型不可解析时，Simple Injector将自动忽略装饰器，这包括泛型类型约束。

换句话说，只需按照以下方式注册装饰器，它就能正常工作：

container.RegisterDecorator(
    typeof(ICommandHandler<>),
    typeof(CleanableCommandHandlerDecorator<>));

Answer 2

我似乎可以使用 DecoratorPredicateContext.ServiceType ：

container.RegisterDecorator(
    typeof(ICommandHandler<>),
    typeof(CleanableCommandHandlerDecorator<>),
    context => {
        var genArg = context.ServiceType.GetGenericArguments()[0];
        return typeof(ICleanableCommand).IsAssignableFrom(genArg);
    });

条件注册基于SimpleInjector中的泛型类型约束的命令装饰器

2 个答案: