c# - Json 对象到 xamarin ListView

Question

我是 xamarin 和 json 的新手，但我已经到了我的 php 服务器通过互联网浏览器得到响应的地步，下面是 json 代码，我知道它是一个对象数组，我为使用 visual studio 选项中的选项的对象(编辑->过去特殊->过去 JSON 作为类)我只找到如何做一个登录系统，但我想做的是在 ListView 上查看数据。对不起我的错英语 。

{
  "error": false,
  "users": [
    {
      "idCustomer": 38,
      "firstname": "Dxxxx",
      "surname": "Cxxxxx",
      "gamertag": "Mxxxx",
      "email": "dpxxxxx@xxxxxl.com",
      "mobile": "012xxxxxx"
    },
    {
      "idCustomer": 36,
      "firstname": "Pxxxxx",
      "surname": "Axxxxxx",
      "gamertag": "Mxxxxxxx",
      "email": "pxxxxxx@xxxxxxx.com",
      "mobile": "0728468780"
    },
    {
      "idCustomer": 37,
      "firstname": "Dxxxxxx",
      "surname": "Dxxxxxx",
      "gamertag": "Axxxxxxxx",
      "email": "txxxxxxx@xxxxxl.com",
      "mobile": "011xxxxxxx"
    },
    {
      "idCustomer": 39,
      "firstname": "Exxxxxx",
      "surname": "Axxxxxxx",
      "gamertag": "Cxxxxxxx",
      "email": "exxxxxx@xxxxxxxx.com",
      "mobile": "012xxxxxxx"
    },
    {
      "idCustomer": 40,
      "firstname": "Cxxxxxx",
      "surname": "Axxxxxxx",
      "gamertag": "Bxxxxxxx",
      "email": "cxxxxxxx.xxxxxxx@xxxxxxx.com",
      "mobile": "012"
    },
    {
      "idCustomer": 41,
      "firstname": "Lenard",
      "surname": "Neemoid",
      "gamertag": "Gamers",
      "email": "email@email.com",
      "mobile": "10214785"
    },
    {
      "idCustomer": 42,
      "firstname": "Lenard",
      "surname": "Nemoi",
      "gamertag": "Spack",
      "email": "email@email.com",
      "mobile": "123456789"
    },
    {
      "idCustomer": 43,
      "firstname": "test",
      "surname": "test",
      "gamertag": "test",
      "email": "test@test.com",
      "mobile": "123456789"
    },
    {
      "idCustomer": 44,
      "firstname": "test",
      "surname": "tees",
      "gamertag": "tes",
      "email": "ttess",
      "mobile": "11222456"
    },
    {
      "idCustomer": 51,
      "firstname": "Dxxxxxxx",
      "surname": "Lxxxxxxx",
      "gamertag": "Sxxxxxxx",
      "email": "Dxxxxxxx.Lxxxxxxx@Sxxxxxxx.com",
      "mobile": "1234567890"
    }
  ]
}

我已经使用了这段代码，但我确实收到了一个错误

namespace LANApp
{
[Activity(Label = "@string/app_name", Theme = "@style/AppTheme", MainLauncher = true)]
public class MainActivity : AppCompatActivity
{
    private ListView list_user;
    private Button bnt_get_data;

    private ArrayAdapter<string> adapter;
    private SynchronizationContext sc;
    protected override void OnCreate(Bundle bundle)
    {
        base.OnCreate(bundle);
        SetContentView(Resource.Layout.Main);

        list_user = FindViewById<ListView>(Resource.Id.list_user);
        bnt_get_data = FindViewById<Button>(Resource.Id.bnt_get_data);

        adapter = new ArrayAdapter<string>(this, Android.Resource.Layout.SimpleDropDownItem1Line);
        adapter.Add("No Data Yet");
        list_user.Adapter = adapter;

        sc = SynchronizationContext.Current;
        bnt_get_data.Click += Bnt_get_data_Click;
    }

    private void Bnt_get_data_Click(object sender, EventArgs e)
    {
        getJSON();
    }


    private async void getJSON()
    {
        adapter.Clear();

        IRestClient client = new RestClient("http://192.xxx.xxx.xxx/Laxxxxxx/public");
        IRestRequest request = new RestRequest("allusers/", Method.GET);

        try
        {
            await Task.Run(() => 
            {
                IRestResponse<List<User>> response = client.Execute<List<User>>(request);
                foreach(var user in response.Data)
                {
                    sc.Post(new SendOrPostCallback(o =>
                    {
                        adapter.Add(o as string);
                        adapter.NotifyDataSetChanged();
                    }), user.gamertag);
                }
            });
        }
        catch(Exception e)
        {
            Console.WriteLine(e.Message);
        }
        return;
    }
}
}

错误是未处理的异常:

Java.Lang.NullPointerException:尝试在空对象引用上调用虚方法“java.lang.String java.lang.Object.toString()”

我的用户类看起来像这样

namespace LANApp.Model
{
public class User
{
    public int idCustomer { get; set; }
    public string firstname { get; set; }
    public string surname { get; set; }
    public string gamertag { get; set; }
    public string email { get; set; }
    public string mobile { get; set; }
}

public class RootObject
{
    public bool error { get; set; }
    public List<User> users { get; set; }
}
}

Answer 1

首先定义一个复杂的对象，您可以将 JSON 反序列化为该对象，它应该看起来像这样:

public class User
{
    public int idCustomer { get; set; }

    public string firstname { get; set; }

    public string surname { get; set; }

    public string gamertag { get; set; }

    public string email { get; set; }

    public string mobile { get; set; }
}

如今，借助 Visual Studio，从 JSON 创建对象非常容易，您可以创建一个新类并复制 JSON 字符串，然后使用:

Edit -> Paste Special -> Paste Json As Classes

这将从您的 JSON 字符串创建一个复杂的对象，就像上面的一样，很简单。

完成后，您可以像这样将 JSON 传递给反序列化器:

using newtonsoft.json

JsonConvert.DeserializeObject<List<User>>(jsonString);

进一步阅读:

• Deserialising an Object

• Creating JSON Objects using C# Guide