将动态内容从一个 Web API 服务发布到另一个答案

【问题标题】：POST dynamic content from one Web API service to another将动态内容从一个 Web API 服务发布到另一个
【发布时间】：2016-03-24 20:11:00
【问题描述】：

我正在使用 ASP.NET Web API 2。从我的应用程序中，我需要将一些动态内容发布到另一个 Web API 服务。目标服务需要这种格式的数据。

public class DataModel
{
    public dynamic Payload { get; set; }
    public string id { get; set; }
    public string key { get; set; }
    public DateTime DateUTC { get; set; }
}

我正在考虑使用这样的东西：

using (var client = new HttpClient())
{                
   client.BaseAddress = new Uri("http://localhost:9000/");
   client.DefaultRequestHeaders.Accept.Clear();
   client.DefaultRequestHeaders.Accept.Add(new  MediaTypeWithQualityHeaderValue("application/json"));
   dynamic payLoad = new ExpandoObject();    
   DataModel model = new DataModel();
   model.Payload = payLoad;    
   var response = await client.PostAsJsonAsync(url, model);    
}

以异步方式将动态信息从一个 Web API 服务发布到另一个的最佳方式是什么？

【问题讨论】：

标签： c# .net asp.net-mvc-4 asp.net-web-api asp.net-web-api2

【解决方案1】：

你就快到了……

根据我可以在您的问题中阅读的内容，以下内容应该适合您。

using (var client = new HttpClient())
{
    var url = new Uri("http://localhost:9000");
    client.DefaultRequestHeaders.Accept.Clear();
    client.DefaultRequestHeaders.Accept.Add(new MediaTypeWithQualityHeaderValue("application/json"));
    dynamic payLoad = new ExpandoObject();

    DataModel model = new DataModel { Payload = payLoad };

    //Model will be serialized automatically.
    var response = await client.PostAsJsonAsync(url, model);

    //It may be a good thing to make sure that your request was handled properly
    response.EnsureSuccessStatusCode();

    //If you need to work with the response, read its content! 
    //Here I implemented the controller so that it sends back what it received
    //ReadAsAsync permits to deserialize the response content
    var responseContent = await response.Content.ReadAsAsync<DataModel>();

    // Do stuff with responseContent...
}

【讨论】：