RestSharp JSON参数发布

Question

我正在尝试对我的MVC 3 API进行非常基本的REST调用,并且我传入的参数不绑定到action方法.

客户

var request = new RestRequest(Method.POST);

request.Resource = "Api/Score";
request.RequestFormat = DataFormat.Json;

request.AddBody(request.JsonSerializer.Serialize(new { A = "foo", B = "bar" }));

RestResponse response = client.Execute(request);
Console.WriteLine(response.Content);

服务器

public class ScoreInputModel
{
   public string A { get; set; }
   public string B { get; set; }
}

// Api/Score
public JsonResult Score(ScoreInputModel input)
{
   // input.A and input.B are empty when called with RestSharp
}

我在这里错过了什么吗？

Answer 1

您不必自己序列化身体.做就是了

request.RequestFormat = DataFormat.Json;
request.AddBody(new { A = "foo", B = "bar" }); // uses JsonSerializer

如果您只想要POST params(它仍然会映射到您的模型并且由于没有序列化到JSON而效率更高),请执行以下操作:

request.AddParameter("A", "foo");
request.AddParameter("B", "bar");

Answer 2

在当前版本的RestSharp(105.2.3.0)中,您可以使用以下命令将JSON对象添加到请求正文:

request.AddJsonBody(new { A = "foo", B = "bar" });

此方法将内容类型设置为application/json,并将对象序列化为JSON字符串.

Answer 3

这对我有用,对我来说这是一个登录请求的帖子:

var client = new RestClient("http://www.example.com/1/2");
var request = new RestRequest();

request.Method = Method.POST;
request.AddHeader("Accept", "application/json");
request.Parameters.Clear();
request.AddParameter("application/json", body , ParameterType.RequestBody);

var response = client.Execute(request);
var content = response.Content; // raw content as string  

身体 :

{
  "userId":"sam@company.com" ,
  "password":"welcome" 
}

Answer 4

希望这会帮助某人。它对我有用-

RestClient client = new RestClient("http://www.example.com/");
RestRequest request = new RestRequest("login", Method.POST);
request.AddHeader("Accept", "application/json");
var body = new
{
    Host = "host_environment",
    Username = "UserID",
    Password = "Password"
};
request.AddJsonBody(body);

var response = client.Execute(request).Content;