将正确的 JSON 作为参数发送到 RPC

Posted 2023-02-16

【中文标题】将正确的 JSON 作为参数发送到 RPC

【英文标题】：Send correct JSON as parameter to RPC

【发布时间】：2020-10-09 20:08:54

【问题描述】：

我需要向第三方 API 创建一个 rpc 并发送以下 JSON

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

我已经创建了一种方法来制作 rcp，但我无法获得要发送的正确 JSON

        public JObject Post()
    
        object[] a_params = new object[]  "\"login\" : \"[e-mail]\"", "\"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;
        
    

使用我的代码，我得到以下 JSON

"jsonrpc":"2.0","id":1,"method":"login.user","params":["\"login\" : \"[e-mail]\"","\"password\": \"AmaYABzP2\""]

据我了解，我的错误是 params 是一个数组（[]）而不是一个对象（）。根据我的方法如何才能得到正确的 json？

【参考方案1】：

我纠正了我的错误。代码应该是

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));