将 Json 读取为对象 MVC

Question

我想从.net MVC上的文件中引用一个json。JSON 位于"内容文件"下。我该怎么做？(这里的例子不起作用。

var url1 = "../../Content/FreeJsonPL.json";
$.getJSON(url1, function (data2) {}

谢谢！！

Answer 1

您需要编写控制器才能从服务器获取文件 下面是一个将返回 Json 结果的示例。在下面的代码中，您需要解析 Json 文件，例如 GetUsersHugeData 函数中的 Usermodel

using System;  
using System.Collections.Generic;  
using System.Text;  
using System.Web.Mvc;  
using System.Web.Script.Serialization;  
using JsonResultDemo.Models;  
namespace JsonResultDemo.Controllers  
{  
    public class JsonDemoController : Controller  
    {          
        /// <summary>  
        /// Get the huge list of Users  
        /// </summary>  
        /// <returns></returns>  
        public JsonResult GetUsersHugeList()  
        {  
            var users = GetUsersHugeData();  
            return Json(users, JsonRequestBehavior.AllowGet);  
        }  
       /// <summary>  
        /// Get the huge list of users  
        /// </summary>  
        /// <returns></returns>  
        private List<UserModel> GetUsersHugeData()  
        {  
            var usersList = new List<UserModel>();  
            UserModel user;  
            for (int i = 1; i < 51000; i++)  
            {  
                user = new UserModel  
                {  
                    UserId = i,  
                    UserName = "User-"+i,  
                    Company = "Company-"+i  
                };  
                usersList.Add(user);  
            }  
            return usersList;  
        }  
        /// <summary>  
        /// Override the Json Result with Max integer JSON lenght  
        /// </summary>  
        /// <param name="data">Data</param>  
        /// <param name="contentType">Content Type</param>  
        /// <param name="contentEncoding">Content Encoding</param>  
        /// <param name="behavior">Behavior</param>  
        /// <returns>As JsonResult</returns>  
        protected override JsonResult Json(object data, string contentType,  
            Encoding contentEncoding, JsonRequestBehavior behavior)  
        {  
            return new JsonResult()  
            {  
                Data = data,  
                ContentType = contentType,  
                ContentEncoding = contentEncoding,  
                JsonRequestBehavior = behavior,  
                MaxJsonLength = Int32.MaxValue  
            };  
        }  
        #endregion  
    }  
} 

然后你需要编写你的获取请求客户端代码

var xhr = new XMLHttpRequest();
xhr.open('GET', "https://domain_name/api/controller_name/method_name", false);
xhr.send();