11

我需要发送这个 HTTP Post 请求：

 POST https://webapi.com/baseurl/login
 Content-Type: application/json

 {"Password":"password",
 "AppVersion":"1",
 "AppComments":"",
 "UserName":"username",
 "AppKey":"dakey" 
  }

就像上面一样，它在 RestClient 和 PostMan 中效果很好。

我需要以编程方式进行此操作，但不确定是否使用

WebClient、HTTPRequest 或 WebRequest 来完成此操作。

问题是如何格式化正文内容并将其与请求一起发送到上面。

这是我使用 WebClient 的示例代码的地方...

  private static void Main(string[] args)
    {
        RunPostAsync();
    } 

    static HttpClient client = new HttpClient();

    private static void RunPostAsync(){

            client.DefaultRequestHeaders.Accept.Clear();
            client.DefaultRequestHeaders.Accept.Add(
                new MediaTypeWithQualityHeaderValue("application/json"));

            Inputs inputs = new Inputs();

            inputs.Password = "pw";
            inputs.AppVersion = "apv";
            inputs.AppComments = "apc";
            inputs.UserName = "user";
            inputs.AppKey = "apk";


            var res = client.PostAsync("https://baseuriplus", new StringContent(JsonConvert.SerializeObject(inputs)));

            try
            {
                res.Result.EnsureSuccessStatusCode();

                Console.WriteLine("Response " + res.Result.Content.ReadAsStringAsync().Result + Environment.NewLine);

            }
            catch (Exception ex)
            {
                Console.WriteLine("Error " + res + " Error " + 
                ex.ToString());
            }

        Console.WriteLine("Response: {0}", result);
    }       

    public class Inputs
    {
        public string Password;
        public string AppVersion;
        public string AppComments;
        public string UserName;
        public string AppKey;
    }

这现在可以工作并使用（200）OK 服务器和响应进行响应

4

5 回答 5

10

你为什么要生成自己的json？

JSONConvert从 JsonNewtonsoft使用。

您的 json 对象字符串值需要" "引号和,

我会使用 http 客户端进行发布，而不是 web 客户端。

using (var client = new HttpClient())
{
   var res = client.PostAsync("YOUR URL", 
     new StringContent(JsonConvert.SerializeObject(
       new { OBJECT DEF HERE },
       Encoding.UTF8, "application/json")
   );

   try
   {
      res.Result.EnsureSuccessStatusCode();
   } 
   catch (Exception e)
   {
     Console.WriteLine(e.ToString());
   }
}
于 2018-05-22T01:23:01.343 回答
1

在发送之前，您没有正确地将值序列化为 JSON。与其尝试自己构建字符串，不如使用 JSON.Net 之类的库。

你可以得到正确的字符串做这样的事情：

var message = JsonConvert.SerializeObject(new {Password = pw, AppVersion = apv, AppComments = acm, UserName = user, AppKey = apk});
Console.WriteLine(message); //Output: {"Password":"password","AppVersion":"10","AppComments":"","UserName":"username","AppKey":"dakey"}
于 2018-05-22T01:25:48.070 回答
0 
            var client = new RestClient("Your URL");
            var request = new RestRequest(Method.POST);
            request.AddHeader("Content-Type", "application/json");
            request.AddHeader("apk-key", apk);

            //Serialize to JSON body.
            JObject jObjectbody = new JObject();
            jObjectbody.Add("employeeName", data.name);
            jObjectbody.Add("designation", data.designation);

            request.AddParameter("application/json", jObjectbody, ParameterType.RequestBody);

            try
            {
                var clientValue= client.Execute<Response>(request);
                return RequestResponse<Response>.Create(ResponseCode.OK, "", clientValue.Data);
            }
            catch (Exception exception)
            {
                throw exception;
            }
于 2020-12-12T13:05:24.970 回答
0

我制作了一个工具来快速轻松地完成它：

Install-Package AdvancedRestHandler

或者

dotnet add package AdvancedRestHandler

AdvancedRestHandler arh = new AdvancedRestHandler("https://webapi.com/baseurl");
var result = await arh.PostDataAsync<MyLoginResponse, MyLoginRequest>("/login", new MyLoginRequest{
  Password = "password",
  AppVersion = "1",
  AppComments = "",
  UserName = "username",
  AppKey = "dakey"
});


public class MyLoginRequest{
  public string Password{get;set;}
  public string AppVersion{get;set;}
  public string AppComments{get;set;}
  public string UserName{get;set;}
  public string AppKey{get;set;}
}

public class MyLoginResponse {
  public string Token{get;set;}
}

额外的：

您可以做的另一件事是使用ArhResponse：

  • 无论哪种方式，在类定义中：
public class MyLoginResponse: ArhResponse 
{
...
}
  • 或者这样，在 API 调用中：
var result = await arh.PostDataAsync<ArhResponse<MyLoginResponse>, MyLoginRequest> (...)

而不是尝试或缓存，而是使用简单的if语句检查您的 API 调用状态：

// check service response status:
if(result.ResponseStatusCode == HttpStatusCode.OK) { /* api receive success response data */ }

// check Exceptions that may occur due to implementation change, or model errors
if(result.Exception!=null) { /* mostly serializer failed due to model mismatch */ }

// have a copy of request and response, in case the service provider need your request response and they think you are hand writing the service and believe you are wrong
_logger.Warning(result.ResponseText);
_logger.Warning(result.RequestText);

// Get deserialized verion of, one of the fallback models, in case the provider uses more than one type of data in same property of the model
var fallbackData = (MyFallbackResponse)result.FallbackModel;

标题可能的问题

在某些情况下，由于生成的标头，服务器不接受 C# 请求HttpClient。

这是因为 HttpClient 默认使用的值application/json; charset=utf-8for Content-Type...

对于仅发送application/json部分Content-Type并忽略该; charset=utf-8部分，您可以执行以下操作：

因为HttpClient您可以通过查看此线程来修复它：How do you set the Content-Type header for an HttpClient request?

至于（AdvancedRestHandler）ARH，由于与某些公司的集成，我已修复它，但我不记得完全......我是通过options类似请求或通过重置header值来完成的。

于 2021-06-02T20:02:29.647 回答
-2

我们将使用 HttpPost 和 HttpClient PostAsync 来解决这个问题。

using System.Net.Http;
    static async Task<string> PostURI(Uri u, HttpContent c)
    {
    var response = string.Empty;
    using (var client = new HttpClient())
    {
    HttpResponseMessage result = await client.PostAsync(u, c);
    if (result.IsSuccessStatusCode)
    {
    response = result.StatusCode.ToString();
    }
    }
    return response;
    }

我们将通过创建一个用于发布的字符串来调用它：

  Uri u = new Uri("http://localhost:31404/Api/Customers");
        var payload = "{\"CustomerId\": 5,\"CustomerName\": \"Pepsi\"}";

        HttpContent c = new StringContent(payload, Encoding.UTF8, "application/json");
        var t = Task.Run(() => PostURI(u, c));
        t.Wait();

        Console.WriteLine(t.Result);
        Console.ReadLine();
于 2019-09-02T11:46:36.407 回答