ExpandoObject對象的JSON序列化

如果：

dynamic expando = new ExpandoObject(); d.SomePRop=SomeValueOrClass;

然后，我們在控制器中：

return new JsonResult(expando);

那么，我們的前臺將會得到：

[{"Key":"SomeProp", "Value": SomeValueOrClass}]

而實際上，我們知道，JSON 格式的內容，應該是這樣的：

{SomeProp: SomeValueOrClass}

于是乎，我們需要一個自定義的序列化器，它應該如下：

public class ExpandoJSONConverter : javaScriptConverter {     public override IEnumerable<Type> SupportedTypes     {         get         {             return new ReadOnlyCollection<Type>(new Type[] { typeof(System.Dynamic.ExpandoObject) });         }     }

    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;     } }

現在，我們的控制器應該像這樣寫：

public ContentResult GetSomeThing(string categores) {     return ControllProctector.Do1(() =>         {

            …             var serializer = new JavaScriptSerializer();             serializer.RegisterConverters(new JavaScriptConverter[] { new ExpandoJSONConverter() });             var json = serializer.Serialize(expando);             return new ContentResult             {                 Content = json,                 ContentType = "application/json"             };         }); }

我們的瀏覽器就能得到正確的 JSON 字符串了。

備注：其它的方法還有

一：

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);

二： JSON.NET

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

var json = JsonConvert.SerializeObject(expando);

三：Content-method:

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

    var json = JsonConvert.SerializeObject(expando);

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

參考：http://stackoverflow.com/questions/5156664/how-to-flatten-an-expandoobject-returned-via-jsonresult-in-asp.net-mvc