ActionResult is an abstract class that an action can return. The helper methods in Controller (eg, Json(), Content(), View(), …) return different concrete classes that inherit ActionResult, including JsonResult. You should declare your action methods as returning ActionResult, so that they have the freedom to return any concrete result class.
I know I’m a bit late on this guys, but I found out why the dynamic solution wasn’t working: JsonResult returns an anonymous object and these are, by default, internal, so they need to be made visible to the tests project. Open your ASP.NET MVC application project and find AssemblyInfo.cs from folder called Properties. Open … Read more
You don’t need to return a JsonResult because its job is to serialize an object into JSON string. You already have the JSON string, so just return it in a ContentResult and specify the correct content type: string json = //get some json from your DB return new ContentResult { Content = json, ContentType = … Read more
I believe the best way to do it, is – as described in your links – to extend ActionResult or extend JsonResult directly. As for the method JsonResult that is not virtual on the controller that’s not true, just choose the right overload. This works well: protected override JsonResult Json(object data, string contentType, Encoding contentEncoding) … Read more
It appears this has been fixed in MVC4. You can do this, which worked well for me: public ActionResult SomeControllerAction() { var jsonResult = Json(veryLargeCollection, JsonRequestBehavior.AllowGet); jsonResult.MaxJsonLength = int.MaxValue; return jsonResult; }
function isJson($string) { json_decode($string); return json_last_error() === JSON_ERROR_NONE; }