ASP.NET MVC: How to create an action filter to output JSON?
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22-08-2019 - |
Question
My second day with ASP.NET MVC and my first request for code on SO (yep, taking a short cut).
I am looking for a way to create a filter that intercepts the current output from an Action and instead outputs JSON (I know of alternate approaches but this is to help me understand filters). I want to ignore any views associated with the action and just grab the ViewData["Output"], convert it to JSON and send it out the client. Blanks to fill:
TestController.cs:
[JSON]
public ActionResult Index()
{
ViewData["Output"] = "This is my output";
return View();
}
JSONFilter.cs:
public override void OnActionExecuting(ActionExecutingContext filterContext)
{
/*
* 1. How to override the View template and set it to null?
* ViewResult { ViewName = "" } does not skip the view (/Test/Index)
*
* 2. Get existing ViewData, convert to JSON and return with appropriate
* custom headers
*/
}
Update: Community answers led to a fuller implementation for a filter for JSON/POX.
Solution
I would suggest that what you really want to do is use the Model rather than arbitrary ViewData
elements and override OnActionExecuted
rather than OnActionExecuting
. That way you simply replace the result with your JsonResult
before it gets executed and thus rendered to the browser.
public class JSONAttribute : ActionFilterAttribute
{
...
public override void OnActionExecuted( ActionExecutedContext filterContext)
{
var result = new JsonResult();
result.Data = ((ViewResult)filterContext.Result).Model;
filterContext.Result = result;
}
...
}
[JSON]public ActionResult Index()
{
ViewData.Model = "This is my output";
return View();
}
OTHER TIPS
You haven't mentioned only returning the JSON conditionally, so if you want the action to return JSON every time, why not use:
public JsonResult Index()
{
var model = new{ foo = "bar" };
return Json(model);
}
maybe this post could help you the right way. The above post is also a method