如何使用异步做一个HTTP GET请求，并等待

Question

我试图创建一个函数，它接受一个url作为参数，并返回从中获得响应，它只是给我一个错误在WebResponse responseObject intialization线

唯一的例外是An exception of type 'System.Net.ProtocolViolationException' occurred in mscorlib.ni.dll but was not handled in user code

public static async Task<string> get(string url) 
{ 
    var request = WebRequest.Create(new Uri(url)) as HttpWebRequest; 
    request.Method = "GET"; 
    request.ContentType = "application/json"; 
    WebResponse responseObject = await Task<WebResponse>.Factory.FromAsync(request.BeginGetResponse, request.EndGetResponse, request); 
    var responseStream = responseObject.GetResponseStream(); 
    var sr = new StreamReader(responseStream); 
    string received = await sr.ReadToEndAsync(); 

    return received; 
} 

Answer 1

不能为GET请求设置ContentType，你是不是将任何数据发送到服务器。

Content-Type =请求主体的MIME类型（与POST和PUT请求一起使用）。

这是您的ProtocolViolationException的来源。

它看起来像你想设置Accept头。

接受=内容类型是可以接受的响应

（按Wikipedia）

试着改变你的代码：以异步方法

request.Accept = "application/json"; 

Answer 2

与HttpClient的尝试它，而不是像这样：

public async Task<List<MyClass>> GetMyClassAsync(
    CancellationToken cancelToken = default(CancellationToken)) 
    { 
     using (HttpClient httpClient = new HttpClient()) 
     { 
      var uri = Util.getServiceUri("myservice"); 
      var response = await httpClient.GetAsync(uri, cancelToken); 
      return (await response.Content.ReadAsAsync<List<MyClass>>()); 
     } 
    }