将正确的JSON作为参数发送到RPC-C#/.NET问题

Send correct JSON as parameter to RPC(将正确的JSON作为参数发送到RPC)

本文介绍了将正确的JSON作为参数发送到RPC的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我需要对第三方API进行RPC并发送以下JSON

{
  "jsonrpc":"2.0",
  "id":"number",
  "method":"login.user",
  "params":{
  "login":"string",
  "password":"string"
  }
}

我已经创建了一个方法来创建RCP，但我无法获得要发送的正确JSON

        public JObject Post()
    {
        object[] a_params = new object[] { ""login" : "test@test.ru"", ""password": "Password""};

        HttpWebRequest webRequest = (HttpWebRequest)WebRequest.Create("https://test.test.ru/v2.0");

        webRequest.ContentType = "application/json; charset=UTF-8";
        webRequest.Method = "POST";

        JObject joe = new JObject();
        joe["jsonrpc"] = "2.0";
        joe["id"] = 1;
        joe["method"] = "login.user";

        if (a_params != null)
        {
            if (a_params.Length > 0)
            {
                JArray props = new JArray();
                foreach (var p in a_params)
                {
                    props.Add(p);
                }
                joe.Add(new JProperty("params", props));
            }
        }

        string s = JsonConvert.SerializeObject(joe);
        // serialize json for the request
        byte[] byteArray = Encoding.UTF8.GetBytes(s);
        webRequest.ContentLength = byteArray.Length;


        WebResponse webResponse = null;
        try
        {
            using (webResponse = webRequest.GetResponse())
            {
                using (Stream str = webResponse.GetResponseStream())
                {
                    using (StreamReader sr = new StreamReader(str))
                    {
                        return JsonConvert.DeserializeObject<JObject>(sr.ReadToEnd());
                    }
                }
            }
        }
        catch (WebException webex)
        {

            using (Stream str = webex.Response.GetResponseStream())
            {
                using (StreamReader sr = new StreamReader(str))
                {
                    var tempRet = JsonConvert.DeserializeObject<JObject>(sr.ReadToEnd());
                    return tempRet;
                }
            }

        }
        catch (Exception)
        {

            throw;
        }
    }

通过我的代码，我得到了以下JSON

{"jsonrpc":"2.0","id":1,"method":"login.user","params":[""login" : "v.ermachenkov@mangazeya.ru"",""password": "AmaYABzP2""]}

据我所知，我的错误是参数是一个数组([])，而不是一个对象({})。根据我的方法，如何才能获得正确的json？

推荐答案

我更正了错误。代码应为

JObject a_params = new JObject { new JProperty("login", "login"), new JProperty("password", "Password") };

        HttpWebRequest webRequest = (HttpWebRequest)WebRequest.Create("https://test.test.ru/v2.0");

        webRequest.ContentType = "application/json; charset=UTF-8";
        webRequest.Method = "POST";

        JObject joe = new JObject();
        joe["jsonrpc"] = "2.0";
        joe["id"] = "1";
        joe["method"] = "login.user";
        joe.Add(new JProperty("params", a_params));

这篇关于将正确的JSON作为参数发送到RPC的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持编程学习网！