WebAPI通过HttpClient访问时如何指定返回格式

[娃都会打酱油了]

WebAPI相关资料好难找。。。。。找资料找了半天都没找到具体的，按照经验去尝试也没尝试出结果来
求哪位大神给个结果，最好能给个实际的响应过程说明

var httpClient = new HttpClient();
string url="http://demo.com/api/demo";
var dic = new Dictionary<string, string>();
//填充Dictionary
HttpContent httpContent = new FormUrlEncodedContent(dic);
//FormUrlEncodedContent默认ContentType 为application/x-www-form-urlencoded
//httpContent 是请求的Content信息，不对应Response信息，修改不正确会导致服务端获取不到Form数据
var taskP = httpClient.PostAsync(url, httpContent );//POST
var taskG = httpClient.GetAsync(url);//GET
//问题在这里，什么时候设定响应的ContentType
HttpResponseMessage  rmP=taskP.Result;
HttpResponseMessage  rmG=taskG.Result;
//HttpResponseMessage  有个Header参数，但貌似不是设定响应类型的
//没搞懂Response响应步骤是如何的，是获取到HttpResponseMessage已经代表整个完整的响应了
//还是通过Content进行读取时才从服务端获取实际的响应数据
//但根据下面的测试设定无效，似乎可以推断在获取HttpResponseMessage时服务器已经给出响应结果了
var responseContent = rmP.Content;
responseContent.Headers.ContentType = MediaTypeHeaderValue.Parse("application/xml");//这步再设定ContentType 已经不起效果
responseContent.ReadAsStringAsync().Result;//这边永远返回的是json格式字符串

[娃都会打酱油了]

尼玛一下午都没找到答案，发完帖子就找到答案了。。。这让我情何以堪。。。。。。。。。

httpClient.DefaultRequestHeaders.Accept.Add(new MediaTypeWithQualityHeaderValue("text/xml“);
httpClient.DefaultRequestHeaders.Accept.Add(new MediaTypeWithQualityHeaderValue("text/json“);

[宝_爸]

可以使用RestSharp来帮助调用web api http://restsharp.org/ 下面是sample code. header中加入accept application/xml应该就可以了。 request.AddHeader("Accept", "application/xml") var client = new RestClient("http://example.com"); // client.Authenticator = new HttpBasicAuthenticator(username, password); var request = new RestRequest("resource/{id}", Method.POST); request.AddParameter("name", "value"); // adds to POST or URL querystring based on Method request.AddUrlSegment("id", "123"); // replaces matching token in request.Resource // easily add HTTP Headers request.AddHeader("header", "value"); // add files to upload (works with compatible verbs) request.AddFile(path); // execute the request RestResponse response = client.Execute(request); var content = response.Content; // raw content as string // or automatically deserialize result // return content type is sniffed but can be explicitly set via RestClient.AddHandler(); RestResponse<Person> response2 = client.Execute<Person>(request); var name = response2.Data.Name; // easy async support client.ExecuteAsync(request, response => { Console.WriteLine(response.Content); }); // async with deserialization var asyncHandle = client.ExecuteAsync<Person>(request, response => { Console.WriteLine(response.Data.Name); }); // abort the request on demand asyncHandle.Abort();

[threenewbee]

很难找么 google webapi return xml 第一条 http://stackoverflow.com/questions/11662028/webapi-to-return-xml 第二个答案 Accept: application/xml 当然，如果你非要坚持用百度，当我没说。

[winnowc]

http://www.asp.net/web-api/overview/advanced/calling-a-web-api-from-a-net-client