[TestFixture]
public class DynamicJson
{
    [Test]
    public void Test()
    {
        dynamic flexible = new ExpandoObject();
        flexible.Int = 3;
        flexible.String = "hi";

        var dictionary = (IDictionary<string, object>)flexible;
        dictionary.Add("Bool", false);

        var serialized = JsonConvert.SerializeObject(dictionary); // {"Int":3,"String":"hi","Bool":false}
    }
}

我发现了一个非常类似于DPeden的解决方案，虽然不需要使用IDictionary，但可以直接从ExpandoObject传递到JSON转换：

dynamic foo = new ExpandoObject();
foo.Bar = "something";
foo.Test = true;
string json = Newtonsoft.Json.JsonConvert.SerializeObject(foo);

并且输出变为：

{ "FirstName":"John", "LastName":"Doe", "Active":true }

你应该使用JavaScriptSerializer。它可以将实际类型序列化为JSON：）
参考：http://msdn.microsoft.com/en-us/library/system.web.script.serialization.javascriptserializer.aspx

var columns = new Dictionary<string, string>
            {
                { "FirstName", "Mathew"},
                { "Surname", "Thompson"},
                { "Gender", "Male"},
                { "SerializeMe", "GoOnThen"}
            };

var jsSerializer = new JavaScriptSerializer();

var serialized = jsSerializer.Serialize(columns);

输出：

{"FirstName":"Mathew","Surname":"Thompson","Gender":"Male","SerializeMe":"GoOnThen"}

使用dynamic和JObject：

dynamic obj = new JObject();
obj.ProductName = "Elbow Grease";
obj.Enabled = true;
obj.StockCount = 9000;

另一种分配属性的方法：

var obj = new JObject();
obj["ProductName"] = "Elbow Grease";
obj["Enabled"] = true;
obj["StockCount"] = 9000;

或者使用JObject.FromObject：

JObject obj = JObject.FromObject(new
{
    ProductName = "Elbow Grease",
    Enabled = true,
    StockCount = 9000
});

它们都产生这样的结果：

Console.WriteLine(obj.ToString());
// {
//   "ProductName": "Elbow Grease",
//   "Enabled": true,
//   "StockCount": 9000
// }

https://www.newtonsoft.com/json/help/html/CreateJsonDynamic.htm