我在我的代码中使用这样的代码，它工作得很好

using System.Web.Script.Serialization;
JavaScriptSerializer oJS = new JavaScriptSerializer();
RootObject oRootObject = new RootObject();
oRootObject = oJS.Deserialize<RootObject>(Your JSon String);

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.

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。

你可以使用Newtonsoft。Json

var jRoot = 
 JsonConvert.DeserializeObject<dynamic>(Encoding.UTF8.GetString(resolvedEvent.Event.Data));

data是我调用core Event得到的响应。

试试这个:

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

在JSON中反序列化。NET可以使用包含在该库中的JObject类来实现动态。我的JSON字符串表示这些类:

public class Foo {
   public int Age {get;set;}
   public Bar Bar {get;set;}
}

public class Bar {
   public DateTime BDay {get;set;}
}

现在我们在不引用上述类的情况下反序列化字符串:

var dyn = JsonConvert.DeserializeObject<JObject>(jsonAsFooString);

JProperty propAge = dyn.Properties().FirstOrDefault(i=>i.Name == "Age");
if(propAge != null) {
    int age = int.Parse(propAge.Value.ToString());
    Console.WriteLine("age=" + age);
}

//or as a one-liner:
int myage = int.Parse(dyn.Properties().First(i=>i.Name == "Age").Value.ToString());

或者如果你想深入一点:

var propBar = dyn.Properties().FirstOrDefault(i=>i.Name == "Bar");
if(propBar != null) {
    JObject o = (JObject)propBar.First();
    var propBDay = o.Properties().FirstOrDefault (i => i.Name=="BDay");
    if(propBDay != null) {
        DateTime bday = DateTime.Parse(propBDay.Value.ToString());
        Console.WriteLine("birthday=" + bday.ToString("MM/dd/yyyy"));
    }
}

//or as a one-liner:
DateTime mybday = DateTime.Parse(((JObject)dyn.Properties().First(i=>i.Name == "Bar").First()).Properties().First(i=>i.Name == "BDay").Value.ToString());

完整的示例请参见文章。