Question

我“创建”了一个非常好的在线工具来为我创建一个C＃类（http://json2csharp.com/），但现在我需要创建一个使用JSON的WebClient并使用它们各自的值填充属性。

实际上我已经尝试创建自己的WebClient，但我不知道如何将它实现为JSON2Csharp生成的代码。

JSON2CSharp代码：

 public class UserModel 
 {
    public int communityvisibilitystate { get; set; }
    public int profilestate { get; set; }
    public string personaname { get; set; }
    public int lastlogoff { get; set; }
    public string profileurl { get; set; }
    public string avatar { get; set; }
    public string avatarmedium { get; set; }
    public string avatarfull { get; set; }
    public int personastate { get; set; }
}

public class Response
{
    public List<UserModel> users { get; set; }
}

public class RootObject
{
    public Response response { get; set; }
}

现在，我尝试使用UserModel的类：

public class UserModel
{
    public UserModel()
    {
        WebClient request = new WebClient();
        var response = request.DownloadString(url);

        JsonConvert.PopulateObject(response, this);
    }

    ...
}

但没有成功（正如预期的那样）。所以我问：我能做些什么来使我的WebClient与JSON2Csharp的代码兼容？

提前致谢。

====更新====

我的观点：

@model DotaMix.Models.UserModel
@{
   ViewBag.Title = @Model.personaname;
}

问题：标题是不可见的。

====更新====

我的完整型号代码：

using System;
using System.Collections.Generic;
using System.ComponentModel.DataAnnotations;
using System.ComponentModel.DataAnnotations.Schema;
using System.Data.Entity;
using System.Globalization;
using System.Web;
using System.Web.Security;
using System.Net;
using System.Configuration;
using System.IO;
using System.Runtime.Serialization.Json;
using System.Text;
using Newtonsoft.Json;
using Newtonsoft.Json.Linq;
using ServiceStack.Text;

namespace Mix.Models
{
    public class UserModel
    {
        public UserModel()
        {
            WebClient request = new WebClient();
            var response = request.DownloadString(url);

            JsonConvert.PopulateObject(response, this);
        }

        public string steamid { get; set; }
        public int communityvisibilitystate { get; set; }
        public int profilestate { get; set; }
        public string personaname { get; set; }
        public int lastlogoff { get; set; }
        public string profileurl { get; set; }
        public string avatar { get; set; }
        public string avatarmedium { get; set; }
        public string avatarfull { get; set; }
        public int personastate { get; set; }
    }

    public class Response
    {
        public List<UserModel> users { get; set; }
    }

    public class RootObject
    {
        public Response response { get; set; }
    }
}

Answer 1

首先，请查看有关如何执行所需操作的完整代码：http://pastebin.com/aQNtjLHs

您获得的JSON可以返回超过1个玩家。如果您确定它只返回1，您可以像这样使用：

控制器：

public ActionResult Index()
{
    WebClient request = new WebClient();
    var url = "http://api.steampowered.com/ISteamUser/GetPlayerSummaries/v0002/?    key=B88B7699C6EACAE138DDB93BED6467EC&steamids=76561198026602449";
    var response = request.DownloadString(url);

    var myObject = new RootObject();

    JsonConvert.PopulateObject(response, myObject);
    //get only the first player     
    return View(myObject.response.players.FirstOrDefault());
}

查看：

@model DotaMix.Models.Player

@Model.personaname

现在，如果那个json真的返回超过1个玩家，你可以这样做：

控制器：

public ActionResult Index()
{

    WebClient request = new WebClient();
    var url = "http://api.steampowered.com/ISteamUser/GetPlayerSummaries/v0002/?key=B88B7699C6EACAE138DDB93BED6467EC&steamids=76561198026602449";
    var response = request.DownloadString(url);
    var myObject = new RootObject();
    JsonConvert.PopulateObject(response, myObject);
    return View(myObject.response);
}

查看：

@model DotaMix.Models.Response

@foreach (var player in Model.players)
{
  <li>@player.personaname</li>
}

Answer 2

我见过你正在使用Newtonsoft程序集。您是否尝试过包含JsonProperty定义？。

[JsonProperty("TheProperty")]

检查此示例类：

namespace Json.CareerJsonTypes
{

    internal class Act13
    {

        [JsonProperty("completed")]
        public bool Completed { get; set; }

        [JsonProperty("completedQuests")]
        public IList<CompletedQuest9> CompletedQuests { get; set; }
    }
}

WebClient使用JSON并发送属性值

2 个答案: