试试这个:

  var units = new { Name = "Phone", Color= "White" };
    var jsonResponse = JsonConvert.DeserializeAnonymousType(json, units);

你想要的DynamicJSONObject对象包含在ASP. web . helpers .dll中。NET Web Pages包，它是WebMatrix的一部分。

JsonFx可以将JSON内容反序列化为动态对象。

序列化动态类型(.NET 4.0的默认值):

var reader = new JsonReader(); var writer = new JsonWriter();

string input = @"{ ""foo"": true, ""array"": [ 42, false, ""Hello!"", null ] }";
dynamic output = reader.Read(input);
Console.WriteLine(output.array[0]); // 42
string json = writer.Write(output);
Console.WriteLine(json); // {"foo":true,"array":[42,false,"Hello!",null]}

用Newtonsoft创建动态对象。Json真的很棒。

//json is your string containing the JSON value
dynamic data = JsonConvert.DeserializeObject<dynamic>(json);

现在您可以像访问普通对象一样访问数据对象。这是我们当前的JSON对象示例:

{ "ID":123,"Name":"Jack","Numbers":[1, 2, 3] }

这是你在反序列化后访问它的方式:

data.ID //Retrieve the int
data.Name //Retrieve the string
data.Numbers[0] //Retrieve the first element in the array

最简单的方法是:

只需包含这个DLL文件。

像这样使用代码:

dynamic json = new JDynamic("{a:'abc'}");
// json.a is a string "abc"

dynamic json = new JDynamic("{a:3.1416}");
// json.a is 3.1416m

dynamic json = new JDynamic("{a:1}");
// json.a is

dynamic json = new JDynamic("[1,2,3]");
/json.Length/json.Count is 3
// And you can use json[0]/ json[2] to get the elements

dynamic json = new JDynamic("{a:[1,2,3]}");
//json.a.Length /json.a.Count is 3.
// And you can use  json.a[0]/ json.a[2] to get the elements

dynamic json = new JDynamic("[{b:1},{c:1}]");
// json.Length/json.Count is 2.
// And you can use the  json[0].b/json[1].c to get the num.

简单的“字符串JSON数据”对象，无需任何第三方DLL文件:

WebClient client = new WebClient();
string getString = client.DownloadString("https://graph.facebook.com/zuck");

JavaScriptSerializer serializer = new JavaScriptSerializer();
dynamic item = serializer.Deserialize<object>(getString);
string name = item["name"];

//note: JavaScriptSerializer in this namespaces
//System.Web.Script.Serialization.JavaScriptSerializer

注意:也可以使用自定义对象。

Personel item = serializer.Deserialize<Personel>(getString);