Вот как я это сделаю
в APi
public ActionResult SomeActionMethod() {
return Json(new {foo="bar", baz="Blech"});
}
Если вы хотите использовать JsonConverter и контролировать способ сериализации данных,
public class JsonNetResult : ActionResult
{
/// <summary>
/// Initializes a new instance of the <see cref="JsonNetResult"/> class.
/// </summary>
public JsonNetResult()
{
}
/// <summary>
/// Gets or sets the content encoding.
/// </summary>
/// <value>The content encoding.</value>
public Encoding ContentEncoding { get; set; }
/// <summary>
/// Gets or sets the type of the content.
/// </summary>
/// <value>The type of the content.</value>
public string ContentType { get; set; }
/// <summary>
/// Gets or sets the data.
/// </summary>
/// <value>The data object.</value>
public object Data { get; set; }
/// <summary>
/// Enables processing of the result of an action method by a custom type that inherits from the <see cref="T:System.Web.Mvc.ActionResult"/> class.
/// </summary>
/// <param name="context">The context in which the result is executed. The context information includes the controller, HTTP content, request context, and route data.</param>
public override void ExecuteResult(ControllerContext context)
{
if (context == null)
{
throw new ArgumentNullException("context");
}
HttpResponseBase response = context.HttpContext.Response;
response.ContentType = !String.IsNullOrWhiteSpace(this.ContentType) ? this.ContentType : "application/json";
if (this.ContentEncoding != null)
{
response.ContentEncoding = this.ContentEncoding;
}
if (this.Data != null)
{
response.Write(JsonConvert.SerializeObject(this.Data));
}
}
}
Blockquote
public class CallbackJsonResult : JsonNetResult
{
/// <summary>
/// Initializes a new instance of the <see cref="CallbackJsonResult"/> class.
/// </summary>
/// <param name="statusCode">The status code.</param>
public CallbackJsonResult(HttpStatusCode statusCode)
{
this.Initialize(statusCode, null, null);
}
/// <summary>
/// Initializes a new instance of the <see cref="CallbackJsonResult"/> class.
/// </summary>
/// <param name="statusCode">The status code.</param>
/// <param name="description">The description.</param>
public CallbackJsonResult(HttpStatusCode statusCode, string description)
{
this.Initialize(statusCode, description, null);
}
/// <summary>
/// Initializes a new instance of the <see cref="CallbackJsonResult"/> class.
/// </summary>
/// <param name="statusCode">The status code.</param>
/// <param name="data">The callback result data.</param>
public CallbackJsonResult(object data, HttpStatusCode statusCode = HttpStatusCode.OK)
{
this.ContentType = null;
this.Initialize(statusCode, null, data);
}
/// <summary>
/// Initializes a new instance of the <see cref="CallbackJsonResult"/> class.
/// </summary>
/// <param name="statusCode">The status code.</param>
/// <param name="description">The description.</param>
/// <param name="data">The callback result data.</param>
public CallbackJsonResult(HttpStatusCode statusCode, string description, object data)
{
this.Initialize(statusCode, description, data);
}
/// <summary>
/// Initializes this instance.
/// </summary>
/// <param name="statusCode">The status code.</param>
/// <param name="description">The description.</param>
/// <param name="data">The callback result data.</param>
private void Initialize(HttpStatusCode statusCode, string description, object data)
{
Data = new JsonData() { Success = statusCode == HttpStatusCode.OK, Status = (int)statusCode, Description = description, Data = data };
}
}
}
затем создайте расширение
/// <summary>
/// return Json Action Result
/// </summary>
/// <typeparam name="T"></typeparam>
/// <param name="o"></param>
/// <param name="JsonFormatting"></param>
/// <returns></returns>
public static CallbackJsonResult ViewResult<T>(this T o)
{
return new CallbackJsonResult(o);
}
Нет APi, просто используйте расширение, которое вы создали
public ActionResult SomeActionMethod() {
return new { error = "", code = (int)HttpStatusCode.OK, data = leads}.ViewResult();
}