Question

Is it possible to control the execution order of custom message handlers?

例如,我可能希望伐木手先执行,因此我总是提出申请。

除了最后增加伐木手,我也看不到如何做到这一点。

config.MessageHandlers.Add(new CustomHandlerOne()); 
config.MessageHandlers.Add(new CustomHandlerTwo());
config.MessageHandlers.Add(new LoggingHandler());

Answer 1

您登记操作员的顺序决定他们何时被点名,但阿利奥斯塔德指出,他们以俄罗斯的代谢模式工作,因此,第一种方式也被称为最后一种。

在即将到来的道路和即将到来的自上而下的道路上以自下而上的方式援引注册的操作员。这就是说,最后的条目首先要求发出一项请求信息,但最后被援引发出即将发出的答复信息。

Answer 2

I am talking here based on the latest bits which are available on Codeplex ASP.NET Web Stack repo.

该命令由用户控制,没有任意命令。让我解释:

Let s say we have two message handlers: MyMessageHandler and MyMessageHandler2. Assuming that we register them as below:

protected void Application_Start(object sender, EventArgs e) {

    RouteConfig.RegisterRoutes(GlobalConfiguration.Configuration.Routes);
    GlobalConfiguration.Configuration.MessageHandlers.Add(new MyMessageHandler());
    GlobalConfiguration.Configuration.MessageHandlers.Add(new MyMessageHandler2());
}

What you expect here is for the MyMessageHandler to run first and MyMessageHandler2 as second one, in other words FIFO.

如果我们看一看框架中的一环,我们将看到<条码>>><条码>方法>。 HttpServer instance isciting CreatePipeline methods of System.Net.Http.HttpClientFactory (以前称为<代码>) HttpPipelineFactory.Create methods as Ali indicated.)CreatePipeline 方法接受两个参数:HttpMessageHandler和IE数<DelegatingHandler>。 <代码>HttpServer.Initialize方法是通过<编码>System.Web.Http.Displer,作为最后<编码>。 HttpMessageHandler within the chain and HttpConfiguration. 信号Handlers for IEvidable<DelegatingHandler> para amount.

<代码>CreatePipeline内发生的情况这种方法非常切合海事组织:

public static HttpMessageHandler CreatePipeline(HttpMessageHandler innerHandler, IEnumerable<DelegatingHandler> handlers)
{
    if (innerHandler == null)
    {
        throw Error.ArgumentNull("innerHandler");
    }

    if (handlers == null)
    {
        return innerHandler;
    }

    // Wire handlers up in reverse order starting with the inner handler
    HttpMessageHandler pipeline = innerHandler;
    IEnumerable<DelegatingHandler> reversedHandlers = handlers.Reverse();
    foreach (DelegatingHandler handler in reversedHandlers)
    {
        if (handler == null)
        {
            throw Error.Argument("handlers", Properties.Resources.DelegatingHandlerArrayContainsNullItem, typeof(DelegatingHandler).Name);
        }

        if (handler.InnerHandler != null)
        {
            throw Error.Argument("handlers", Properties.Resources.DelegatingHandlerArrayHasNonNullInnerHandler, typeof(DelegatingHandler).Name, "InnerHandler", handler.GetType().Name);
        }

        handler.InnerHandler = pipeline;
        pipeline = handler;
    }

    return pipeline;
}

如你所知,电文手令被推翻,。 Matryoshka doll 这里虽已建立,但需谨慎:确保<代码>。 HttpControllerDispader 是链条内最后的电传手。

至于两个问题,这实际上并不完全。贺电手得双倍,而你提供的持续方法则会是这样。你们有责任做到这一点。如果你提供反馈(换言之,继续),你的信息处理员将被要求向客户提供你可以提供的回复信息。

例如,请假设以下两点是上文所述的信息处理器:

public class MyMessageHandler : DelegatingHandler {

    protected override Task<HttpResponseMessage> SendAsync(HttpRequestMessage request, System.Threading.CancellationToken cancellationToken) {

        //inspect request here

        return base.SendAsync(request, cancellationToken).ContinueWith(task => {

            //inspect the generated response
            var response = task.Result;

            return response;
        });
    }
}

这是另一个:

public class MyMessageHandler2 : DelegatingHandler {

    protected override Task<HttpResponseMessage> SendAsync(HttpRequestMessage request, System.Threading.CancellationToken cancellationToken) {

        //inspect request here

        return base.SendAsync(request, cancellationToken).ContinueWith(task => {

            //inspect the generated response
            var response = task.Result;

            return response;
        });
    }
}

随着我们的持续,我们的信息处理员将回到劳工组织命令的客户手中。因此,<代码>内的延续方法 MyMessageHandler2将是背路上首先要援引的,是内的。 MyMessageHandler将是第二个。

Answer 3

No - AFAIK.

它是俄罗斯的多拉模式,在最后一次工作之前,有一位手在另一边。内部栏目<代码> HttpPipelineFactory (在排放时,可以认为来源代码):

    public static HttpMessageHandler Create(IEnumerable<DelegatingHandler> handlers, HttpMessageHandler innerChannel)
    {
        if (innerChannel == null)
        {
            throw Error.ArgumentNull("innerChannel");
        }

        if (handlers == null)
        {
            return innerChannel;
        }

        // Wire handlers up
        HttpMessageHandler pipeline = innerChannel;
        foreach (DelegatingHandler handler in handlers)
        {
            if (handler == null)
            {
                throw Error.Argument("handlers", SRResources.DelegatingHandlerArrayContainsNullItem, typeof(DelegatingHandler).Name);
            }

            if (handler.InnerHandler != null)
            {
                throw Error.Argument("handlers", SRResources.DelegatingHandlerArrayHasNonNullInnerHandler, typeof(DelegatingHandler).Name, "InnerHandler", handler.GetType().Name);
            }

            handler.InnerHandler = pipeline;
            pipeline = handler;
        }

        return pipeline;
    }

因此,《守则》是一份名单,然后将其变为。 Russian Doll。

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