使用xamarin for Android中的对象从Json对象创建对象列表

Question

我正在使用JSON.net，我正试图从JSON中获得一个Expense对象列表，如下所示：

{"ApplicationUser": null, "Currency": 1, "Description": "Moj pierwszy portfel", "Expenses": [{"Date": "2015-10-01T00:00:00", "Description": "Opis", "ExpenseType": {"Id": 1, "IsExpense": true, "Name": "Jedzenie"}, "ExpenseTypeId": 1, "Id": 1, "Name": "Biedronka", "Value": 40.00, "WalletId": 1}], "Id": 1, "Name": "Moj portfel", "Saldo": 100.12, "UserId": "f9b94a9a-44b3-4987-8711-6c1b73a5cb0e"}

api url:http://homecalc.azurewebsites.net/api/wallets/2

我有从JSON2C#构建的类

它们看起来像这个

public class ExpenseType
{
    public int Id { get; set; }
    public bool IsExpense { get; set; }
    public string Name { get; set; }
}
public class Expense
{
    public ExpenseType ExpenseType { get; set; }
    public int Id { get; set; }
    public string Name { get; set; }
    public double Value { get; set; }
    public string Description { get; set; }
    public string Date { get; set; }
    public int ExpenseTypeId { get; set; }
    public int WalletId { get; set; }
}
public class RootObject
{
    public List<Expense> Expenses { get; set; }
    public int Id { get; set; }
    public string Name { get; set; }
    public string UserId { get; set; }
    public double Saldo { get; set; }
    public int Currency { get; set; }
    public string Description { get; set; }
    public object ApplicationUser { get; set; }
}

Getting JSONValue声明如下：

private async Task<JsonValue> GetApi(string url)
    {
        HttpWebRequest request = (HttpWebRequest) HttpWebRequest.Create(new Uri(url));
        request.ContentType = "application/json";
        request.Method = "GET";
        using (WebResponse response = await request.GetResponseAsync())
        {
            using (Stream stream = response.GetResponseStream())
            {
                JsonValue jsconDoc = await Task.Run(() => JsonObject.Load(stream));
                Console.Out.WriteLine("Response {0}", stream);
                return jsconDoc;
            }
        }
    }

我尝试过反序列化

var my_array = JsonConvert.DeserializeObject<List<RootObject>>(json);

并将其解析为jsonObject或jsonArray

var obj = JsonObject.Parse(json);

但它总是以破坏应用程序而告终。我也尝试使用json["Expenses"]只获取一组费用，但当有两个费用时，它会将所有费用作为一个费用返回。有人能解释一下如何从中获得支出对象列表吗？

我把代码改成这样：

 private async Task<string> GetApi(string url)
    {
        HttpWebRequest request = (HttpWebRequest) HttpWebRequest.Create(new Uri(url));
        request.ContentType = "application/json";
        request.Method = "GET";
        using (WebResponse response = await request.GetResponseAsync())
        {
            using (Stream stream = response.GetResponseStream())
            {
                JsonValue jsconDoc = await Task.Run(() => JsonObject.Load(stream));
                Console.Out.WriteLine("Response {0}", stream);
                StreamReader read = new StreamReader(stream,Encoding.UTF8);
                string json = read.ReadToEnd();
                return json;
            }
        }
    }
private void ParseAndDisplay(string json)
    {
        //TextView WalletName = FindViewById<TextView>(Resource.Id.PortfelName);
        //WalletName.Text = json["Name"];
        //Console.WriteLine("Exp {0}",json["Expenses"]);
        //JsonArray jsonnArray = new JsonArray(json["Expenses"]);
        try
        {
            var result = JsonConvert.DeserializeObject<RootObject>(json);
            Console.WriteLine("Nazwa: {0}",result.Name);
        }
        catch (Exception e)
        {
            Console.WriteLine("Błąd: {0}",e);
        }
    }

当我尝试运行这些方法时，它会返回一个异常：

System.NullReferenceException: Object reference not set to an instance of an object

Answer 1

如果不深入了解细节，使用System.Net.Http.HttpClient和类似Newtonsoft.JSON或ServiceStack.Text的JSON库，这将相当简单。使用Newtonsoft.Json，你可能正在寻找这样的东西：

using System.Net.Http;
using Newtonsoft.Json;
...
using (var client = new HttpClient())
{
    ...
    var response = await client.GetAsync(url);
    response.EnsureSuccessStatusCode();
    string json = await response.Content.ReadAsStringAsync();
    var result = JsonConvert.DeserializeObject<RootObject>(json);
    ...
}