如何在asp.net mvc中展平通过JsonResult返回的ExpandoObject？

Question

我非常喜欢在运行ExpandoObject时编译服务器端动态对象,但是在JSON序列化期间我遇到了麻烦.首先,我实例化对象:

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

到现在为止还挺好.在我的MVC控制器中,我想将其作为JsonResult发送,所以我这样做:

return new JsonResult(expando);

这将JSON序列化到下面,由浏览器使用:

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

但是,我真正喜欢的是看到这个:

{SomeProp: SomeValueOrClass}

我知道我可以实现这个,如果我使用dynamic而不是ExpandoObject- JsonResult能够将dynamic属性和值序列化为单个对象(没有Key或Value业务),但我需要使用的原因ExpandoObject是因为我不知道所有的直到运行时我想要的对象属性,据我所知,我不能动态地添加属性dynamic而不使用ExpandoObject.

我可能不得不在我的javascript中筛选"关键","价值"业务,但我希望在将其发送给客户之前解决这个问题.谢谢你的帮助!

Answer 1

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");
    }
}

Answer 2

您还可以创建一个特殊的JSONConverter,它只适用于ExpandoObject,然后在JavaScriptSerializer的实例中注册它.这样你就可以序列化expando的数组,expando对象的组合和...直到找到另一种没有正确序列化的对象("你想要的方式"),然后你创建另一个转换器,或者添加另一个类型到这个.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);

Answer 3

以下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);

成本是您在序列化之前制作数据的副本.

Answer 4

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.

然后JsonResult看起来像这样:

return JsonResult(expando.Flatten());

这将返回给浏览器:

"{SomeProp: SomeValueOrClass}"

我可以在javascript中使用它(在这里引用):

var obj = JSON.parse(myJsonString);

我希望这有帮助!

Answer 5

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 "}