使用 JSON.NET，您可以调用 SerializeObject 来“展平” expando 对象：

dynamic expando = new ExpandoObject();
expando.name = "John Smith";
expando.age = 30;

var json = JsonConvert.SerializeObject(expando);

将输出：

{"name":"John Smith","age":30}

在 ASP.NET MVC 控制器的上下文中，可以使用 Content-method 返回结果：

public class JsonController : Controller
{
    public ActionResult Data()
    {
        dynamic expando = new ExpandoObject();
        expando.name = "John Smith";
        expando.age = 30;

        var json = JsonConvert.SerializeObject(expando);

        return Content(json, "application/json");
    }
}

您还可以制作一个仅适用于 ExpandoObject 的特殊 JSONConverter，然后将其注册到 JavaScriptSerializer 的实例中。通过这种方式，您可以序列化 expando 数组、expando 对象的组合以及......直到您找到另一种没有正确序列化的对象（“你想要的方式”），然后制作另一个转换器，或添加另一种类型到这个。希望这可以帮助。

using System.Web.Script.Serialization;    
public class ExpandoJSONConverter : JavaScriptConverter
{
    public override object Deserialize(IDictionary<string, object> dictionary, Type type, JavaScriptSerializer serializer)
    {
        throw new NotImplementedException();
    }
    public override IDictionary<string, object> Serialize(object obj, JavaScriptSerializer serializer)
    {         
        var result = new Dictionary<string, object>();
        var dictionary = obj as IDictionary<string, object>;
        foreach (var item in dictionary)
            result.Add(item.Key, item.Value);
        return result;
    }
    public override IEnumerable<Type> SupportedTypes
    {
        get 
        { 
              return new ReadOnlyCollection<Type>(new Type[] { typeof(System.Dynamic.ExpandoObject) });
        }
    }
}

使用转换器

var serializer = new JavaScriptSerializer(); 
serializer.RegisterConverters(new JavaScriptConverter[] { new ExpandoJSONConverter()});
var json = serializer.Serialize(obj);

这是我为实现您所描述的行为所做的工作：

dynamic expando = new ExpandoObject();
expando.Blah = 42;
expando.Foo = "test";
...

var d = expando as IDictionary<string, object>;
d.Add("SomeProp", SomeValueOrClass);

// After you've added the properties you would like.
d = d.ToDictionary(x => x.Key, x => x.Value);
return new JsonResult(d);

代价是您在序列化之前制作了数据的副本。

我通过编写一个将 ExpandoObject 转换为 JSON 字符串的扩展方法解决了这个问题：

public static string Flatten(this ExpandoObject expando)
{
    StringBuilder sb = new StringBuilder();
    List<string> contents = new List<string>();
    var d = expando as IDictionary<string, object>;
    sb.Append("{");

    foreach (KeyValuePair<string, object> kvp in d) {
        contents.Add(String.Format("{0}: {1}", kvp.Key,
           JsonConvert.SerializeObject(kvp.Value)));
    }
    sb.Append(String.Join(",", contents.ToArray()));

    sb.Append("}");

    return sb.ToString();
}

这使用了优秀的Newtonsoft库。

JsonResult 然后看起来像这样：

return JsonResult(expando.Flatten());

这将返回给浏览器：

"{SomeProp: SomeValueOrClass}"

我可以通过这样做（在此处引用）在javascript中使用它：

var obj = JSON.parse(myJsonString);

我希望这有帮助！

我能够使用JsonFx解决同样的问题。

        dynamic person = new System.Dynamic.ExpandoObject();
        person.FirstName  = "John";
        person.LastName   = "Doe";
        person.Address    = "1234 Home St";
        person.City       = "Home Town";
        person.State      = "CA";
        person.Zip        = "12345";

        var writer = new JsonFx.Json.JsonWriter();
        return writer.Write(person);

输出：

{ "FirstName": "John", "LastName": "Doe", "Address": "1234 Home St", "City": "Home Town", "State": "CA", "Zip": "12345 " }