我以前没有用JSON做过任何事情...所以我可能只是错过了一步。
下面是我要反序列化的 JSON 示例:
{"item":{"icon":"http://services.runescape.com/m=itemdb_rs/4765_obj_sprite.gif?id=4798","icon_large":"http://services.runescape.com/m=itemdb_rs/4765_obj_big.gif?id=4798","id":4798,"type":"Ammo","typeIcon":"http://www.runescape.com/img/categories/Ammo","name":"Adamant brutal","description":"Blunt adamantite arrow...ouch","current":{"trend":"neutral","price":237},"today":{"trend":"neutral","price":0},"members":"true","day30":{"trend":"positive","change":"+1.0%"},"day90":{"trend":"negative","change":"-0.0%"},"day180":{"trend":"positive","change":"+0.0%"}}}
我把它放到"Json 2 C#"中,最终创建了这个新的.cs文件:
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace RSTool.Models
{
public class Current
{
public string trend { get; set; }
public int price { get; set; }
}
public class Today
{
public string trend { get; set; }
public int price { get; set; }
}
public class Day30
{
public string trend { get; set; }
public string change { get; set; }
}
public class Day90
{
public string trend { get; set; }
public string change { get; set; }
}
public class Day180
{
public string trend { get; set; }
public string change { get; set; }
}
public class Item
{
public string icon { get; set; }
public string icon_large { get; set; }
public int id { get; set; }
public string type { get; set; }
public string typeIcon { get; set; }
public string name { get; set; }
public string description { get; set; }
public Current current { get; set; }
public Today today { get; set; }
public string members { get; set; }
public Day30 day30 { get; set; }
public Day90 day90 { get; set; }
public Day180 day180 { get; set; }
}
public class RootObject
{
public Item item { get; set; }
}
}
所以,我有课程。我可以从其位置以字符串的形式检索 JSON,但我真的不知道如何反序列化它......我已经安装了Newtonsoft.Json,并尝试使用PopulateObject和解序列化程序,但运气不佳...
假设我的 JSON 存储为名为"json"的字符串,例如,我将如何存储该查询然后检索项目名称?
使用:
var deserialized = JsonConvert.DeserializeObject<RootObject>(json);
我刚刚根据您提供的代码成功测试了这一点。
然后,您可以像往常一样访问对象的属性:
MessageBox.Show(deserialized.item.name);