当数组表示为带有"item#XXX"的对象时,读取JSON.特性 [英] Read JSON when array represented as objects with "item#XXX" properties

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本文介绍了当数组表示为带有"item#XXX"的对象时,读取JSON.特性的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我有这个JSON文本,我不知道如何解析"items"属性来填充项目列表

I have this JSON text and I can't figure out how to parse the "items" property to fill up a list of items

{
    "response": {
        "success": 1,
        "current_time": 1445015502,
        "items": {
            "item1": {
                "property1": 1,
                "property2": "test",
                "property3": 4.3
            },
            "item2": {
                "property1": 5,
                "property2": "test2",
                "property3": 7.8
            }
        }
    }
}

那是我的课程吗:

public class Item
{
    public int property1 { get; set; }
    public string property2 { get; set; }
    public double property3 { get; set; }
}

public class Response
{
    public int success { get; set; }
    public string message { get; set; }
    public int current_time { get; set; }
    public List<Item> items { get; set; }
}

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

另外,不,这不是错误. JSON文本中没有[]. 另外,JSON中的项目数是不确定的.

Also, no, it's not a error. There is no [ nor ] in the JSON text. Also, the number of items in the JSON is undefined.

推荐答案

由于 Json.NET,这很容易实现动态:

private RootObject Parse(string jsonString)
{
   dynamic jsonObject = JsonConvert.DeserializeObject(jsonString);
   RootObject parsed = new RootObject()
   {
        response = new Response()
        {
              success = jsonObject.response.success,
              current_time = jsonObject.response.current_time,
              message = jsonObject.response.message,
              items = ParseItems(jsonObject.response.items)
        }  
   };
   return parsed;
}

private List<Item> ParseItems(dynamic items)
{
     List<Item> itemList = new List<Item>();
     foreach (var item in items)
     {
         itemList.Add(new Item()
         {
              property1 = item.Value.property1,
              property2 = item.Value.property2,
              property3 = item.Value.property3
         });
     }
     return itemList;
}

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