как десериализовать список json без использования модели в сервисе?

Question

вот код, который может хорошо работать при использовании модели в сервисе:

Модели:

using System;
using System.Collections.Generic;
using System.Linq;
using System.Web;

namespace listTest.Models
{
    public class Employee
    {
        public int Id { get; set; }
        public string Name { get; set; }
        public string Age { get; set; }
        public string Location { get; set; }
        public string Sex { get; set; }

    }
}

ViewModel:

using listTest.Models;
using System;
using System.Collections.Generic;
using System.Linq;
using System.Web;

namespace listTest.ViewModels
{
    public class ViewModel
    {
        public List<Employee> EmployeeDataList { get; set; }
    }
}

сервис:

using listTest.Models;
using System;
using System.Collections.Generic;
using System.Configuration;
using System.Data.SqlClient;
using System.Linq;
using System.Web;
using System.Data;
using Newtonsoft.Json;


namespace listTest.Service
{
    public class DBService
    {
        private readonly static string cnstr = ConfigurationManager.ConnectionStrings["connection1"].ConnectionString;
       SqlConnection conn = new SqlConnection(cnstr);

        public List<Employee> getData()
        {
            List<Employee> DataList = new List<Employee>();

            string sql = @"select * from Employee";

            try
            {
                conn.Open();
                SqlCommand cmd = new SqlCommand(sql, conn);
                SqlDataReader dr = cmd.ExecuteReader();
                while (dr.Read())
                {
                    Employee emp = new Employee();
                    emp.Id = Convert.ToInt32(dr["Id"]);
                    emp.Name = dr["Name"].ToString();
                    emp.Age = dr["Age"].ToString();
                    emp.Location = dr["Location"].ToString();
                    emp.Sex = dr["Sex"].ToString();
                    DataList.Add(emp);                   
                }
            }
            catch (Exception e)
            {
                throw new Exception(e.Message.ToString());
            }
            finally
            {
                conn.Close();
            }
            return DataList;            
        }

    }
}

контроллер:

using System;
using System.Collections.Generic;
using System.Linq;
using System.Web;
using System.Web.Mvc;
using listTest.Service;
using listTest.ViewModels;
using Newtonsoft.Json;
using listTest.Models;
using Newtonsoft;
using Newtonsoft.Json.Linq;

namespace listTest.Controllers
{
    public class HomeController : Controller
    {
        public ActionResult Index()
        {
            return View();
        }
        public ActionResult EmpList()
        {
            DBService service = new DBService();
            ViewModel data = new ViewModel();
            data.EmployeeDataList=service.getData();

            return View(data);;
        }
    }
}

но теперь я сообщаю, что не могу использовать модель в Сервисе, поэтому я изменил код следующим образом:

Сервис

using listTest.Models;
using System;
using System.Collections.Generic;
using System.Configuration;
using System.Data.SqlClient;
using System.Linq;
using System.Web;
using System.Data;
using Newtonsoft.Json;


namespace listTest.Service
{
    public class DBService
    {
        private readonly static string cnstr = ConfigurationManager.ConnectionStrings["connection1"].ConnectionString;
       SqlConnection conn = new SqlConnection(cnstr);

        //public List<Employee> getData()
        public string getData()
        {
            //List<Employee> DataList = new List<Employee>();
            List<string> DataList = new List<string>();

            string sql = @"select * from Employee";

            try
            {
                conn.Open();
                SqlCommand cmd = new SqlCommand(sql, conn);
                SqlDataReader dr = cmd.ExecuteReader();
                while (dr.Read())
                {
                    //Employee emp = new Employee();
                    //emp.Id = Convert.ToInt32(dr["Id"]);
                    //emp.Name = dr["Name"].ToString();
                    //emp.Age = dr["Age"].ToString();
                    //emp.Location = dr["Location"].ToString();
                    //emp.Sex = dr["Sex"].ToString();
                    //DataList.Add(emp);

                    string Id = dr["Id"].ToString();
                    string Name = dr["Name"].ToString();
                    string Age = dr["Age"].ToString();
                    string Location = dr["Location"].ToString();
                    string Sex = dr["Sex"].ToString();
                    string jsonList = "{"+"Id"+":"+Id+","+"Name"+":"+Name+","+"Age"+":"+Age+","+"Location"+":"+Location+","+"Sex"+":"+Sex+"}";
                    DataList.Add(jsonList);

                }
            }
            catch (Exception e)
            {
                throw new Exception(e.Message.ToString());
            }
            finally
            {
                conn.Close();
            }
            //return DataList;
            return JsonConvert.SerializeObject(DataList);
        }

    }
}

контроллер:

using System;
using System.Collections.Generic;
using System.Linq;
using System.Web;
using System.Web.Mvc;
using listTest.Service;
using listTest.ViewModels;
using Newtonsoft.Json;
using listTest.Models;
using Newtonsoft;
using Newtonsoft.Json.Linq;

namespace listTest.Controllers
{
    public class HomeController : Controller
    {
        public ActionResult Index()
        {
            return View();
        }
        public ActionResult EmpList()
        {
            DBService service = new DBService();
            ViewModel data = new ViewModel();
            //data.EmployeeDataList=service.getData();



           //don't know how to do 
            data.EmployeeDataList = JsonConvert.DeserializeObject<Employee>(service.getData());


            return View(data);
        }
    }
}

что я хочу - десериализовать список и сохранить его в данных. EmployeeDataList
как мне поступить? Пожалуйста, помогите мне

Служба обновления:

using listTest.Models;
using System;
using System.Collections.Generic;
using System.Configuration;
using System.Data.SqlClient;
using System.Linq;
using System.Web;
using System.Data;
using Newtonsoft.Json;
using System.Dynamic;

namespace listTest.Service
{
    public class DBService
    {
        private readonly static string cnstr = ConfigurationManager.ConnectionStrings["connection1"].ConnectionString;
       SqlConnection conn = new SqlConnection(cnstr);

        //public List<Employee> getData()
        public string getData()
        //public List<string> getData()
        {

            List<string> DataList = new List<string>();
            //List<Employee> DataList = new List<Employee>();
            //List<dynamic> DataList = new List<dynamic>();

            string sql = @"select * from Employee";

            try
            {
                conn.Open();
                SqlCommand cmd = new SqlCommand(sql, conn);
                SqlDataReader dr = cmd.ExecuteReader();
                while (dr.Read())
                {
                    //Employee emp = new Employee();
                    //emp.Id = Convert.ToInt32(dr["Id"]);
                    //emp.Name = dr["Name"].ToString();
                    //emp.Age = dr["Age"].ToString();
                    //emp.Location = dr["Location"].ToString();
                    //emp.Sex = dr["Sex"].ToString();
                    //DataList.Add(emp);

                    dynamic myObj = new ExpandoObject();

                    myObj.Id = dr["Id"].ToString();
                    myObj.Name = dr["Name"].ToString();
                    myObj.Age = dr["Age"].ToString();
                    myObj.Location = dr["Location"].ToString();
                    myObj.Sex = dr["Sex"].ToString();

                    DataList.Add(myObj);

                    //string Id = dr["Id"].ToString();
                    //string Name = dr["Name"].ToString();
                    //string Age = dr["Age"].ToString();
                    //string Location = dr["Location"].ToString();
                    //string Sex = dr["Sex"].ToString();
                    //string jsonList = "{" + "Id" + ":" + Id + "," + "Name" + ":" + Name + "," + "Age" + ":" + Age + "," + "Location" + ":" + Location + "," + "Sex" + ":" + Sex + "}";
                    //DataList.Add(jsonList);

                }
            }
            catch (Exception e)
            {
                throw new Exception(e.Message.ToString());
            }
            finally
            {
                conn.Close();
            }
            //return DataList;
            return JsonConvert.SerializeObject(DataList);
        }

    }
}

Answer 1

Если вам не разрешено использовать Модели, вы можете использовать встроенный ExpandoObject, чтобы создать объект и затем сохранить его в списке. Как только вы закончите, сохраните его там, где вам нужно.

Примеры:

Сериализация

ExpandoObject

try
{
    conn.Open();
    SqlCommand cmd = new SqlCommand(sql, conn);
    SqlDataReader dr = cmd.ExecuteReader();

    // Use a list of dynamic variables instead of string.
    var DataList = new List<dynamic>();

    while (dr.Read())
    {
        dynamic myObj = new ExpandoObject();

        myObj.Id = dr["Id"].ToString();
        myObj.Name = dr["Name"].ToString();
        myObj.Age = dr["Age"].ToString();
        myObj.Location = dr["Location"].ToString();
        myObj.Sex = dr["Sex"].ToString();

        DataList.Add(myObj); // Add ExpandoObject as it is without conversion.
    }
}

Словарь

Если вам неудобно использовать ExpandoObject, возможно, вам подойдет словарь.

try
{
    conn.Open();
    SqlCommand cmd = new SqlCommand(sql, conn);
    SqlDataReader dr = cmd.ExecuteReader();

    var DataList = new List<dynamic>();
    while (dr.Read())
    {

        var myObj = new Dictionary<string, string>();

        myObj["Id"] = dr["Id"].ToString();
        myObj["Name"] = dr["Name"].ToString();
        myObj["Age"] = dr["Age"].ToString();
        myObj["Location"] = dr["Location"].ToString();
        myObj["Sex"] = dr["Sex"].ToString();

        DataList.Add(myObj); // Add dictionary as an object to List.
    }
}

---

Десериализация

Вы можете использовать следующее для десериализации объектов и чтения их свойств.

ExpandoObject или Dictionary

public ActionResult EmpList()
{
    DBService service = new DBService();
    ViewModel data = new ViewModel();

    // Just to make sure its a string
    string strdata = service.getData();

    data.EmployeeDataList = JsonConvert.DeserializeObject<List<Employee>>(strData).ToList();

    return View(data);
}

---

Обновление

Тестовый набор с использованием циклов.

var DataList = new List<dynamic>();
List<Employee> EmployeeDataList = new List<Employee>();
for (int i = 0; i < 4; i++)
{
    var myObj = new Dictionary<string, string>();

    myObj["Id"] = i.ToString();
    myObj["Name"] = "Name" + i.ToString();
    myObj["Age"] = "Age" + i.ToString();
    myObj["Location"] = "Location" + i.ToString();
    myObj["Sex"] = "Sex" + i.ToString();


    DataList.Add(myObj); // Add Dictionary to List.
}

for (int i = 0; i < 4; i++)
{
    dynamic myObj = new ExpandoObject();

    myObj.Id = i.ToString();
    myObj.Name = "Name" + i.ToString();
    myObj.Age = "Age" + i.ToString();
    myObj.Location = "Location" + i.ToString();
    myObj.Sex = "Sex" + i.ToString();

    DataList.Add(myObj); // Add Expando Object to List
}

// Represents the data you will recieve from service.getData();
string strData = JsonConvert.SerializeObject(DataList);

// represents the list of Employee.
EmployeeDataList = JsonConvert.DeserializeObject<List<Employee>>(strData).ToList();