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Asp.net core Cleanest way to return View or Json/XML

In asp.net core I would like to set up my API controller to do the following:
by default return View(model);
/api/id.json to return model; as json
/api/id.xml to return model; as xml
The second two can be achieved by using the [FormatFilter] see here
[FormatFilter]
public class ProductsController
{
[Route("[controller]/[action]/{id}.{format?}")]
public Product GetById(int id)
However this requires the method to return an object and not a View(object). Is there anyway to cleanly support also returning Views?

You cannot do both in the same action. However, you can factor out the common functionality into a private method and then implement two actions with minimal code duplication:
[Route("[controller]")]
[FormatFilter]
public class ProductsController : Controller
{
private Product GetByIdCore(int id)
{
// common code here, return product
}
[HttpGet("[action]/{id}")]
[ActionName("GetById")]
public IActionResult GetByIdView(int id) => View(GetByIdCore(id));
[HttpGet("[action]/{id}.{format}")]
public Product GetById(int id) => GetByIdCore(id);
}
It's necessary to use different action names here, because the method signatures cannot differ merely on return type. However, the [ActionName] attribute can be used as above to make them appear to have the same name for the purposes of URL generation and such.

You can actually achieve this just using the one action. Here's an example of how I got it to work:
[FormatFilter]
public class ProductsController : Controller
{
[Route("[controller]/[action]/{id}.{format?}")]
public IActionResult GetById(int id, string format)
{
var yourModel = ...;
if (string.IsNullOrWhiteSpace(format))
return View(yourModel);
return Ok(yourModel);
}
By using IActionResult as the return type, you can return either a ViewResult or an OkObjectResult. You can get access to the format value by taking it as a parameter in your action, check if it's empty and then react accordingly.
I also added Controller as the base class in order to access the convenience methods for creating the relevant results (View(...) and Ok(...)).

If you're going to be using this pattern a lot, to keep your controllers as clean as possible, you could create a base class that exposed a "FormatOrView" method:
[FormatFilter]
public abstract class FormatController : Controller
{
protected ActionResult FormatOrView(object model)
{
var filter = HttpContext.RequestServices.GetRequiredService<FormatFilter>();
if (filter.GetFormat(ControllerContext) == null)
{
return View(model);
}
else
{
return new ObjectResult(model);
}
}
}
And then your controller can inherit from this and use the FormatOrView method
public class ProductsController : FormatController
{
[Route("[controller]/[action]/{id}.{format?}")]
public ActionResult GetById(int id)
{
var product = new { Id = id };
return FormatOrView(product);
}
}
Edit to list final accepted answer by GreyCloud: Here is a generic slightly simplified method you can put into a controller (or make an extension method or put into an abstract base class as above). Note the ?. in case the service is not defined for some reason.
private ActionResult<T> FormatOrView<T>(T model) {
return HttpContext.RequestServices.GetRequiredService<FormatFilter>()?.GetFormat(ControllerContext) == null
? View(model)
: new ActionResult<T>(model);
}

The FormatFilter is part of the content negotiation of your app, in AspNetCore, you have the control to handle your input or output formatters also on the ConfigureServices where you have more control, even you can add more media types there
public void ConfigureServices(IServiceCollection services)
{
services.AddMvc(options =>
{
options .OutputFormatters.Add(new XmlDataContractSerializerOutputFormatter());
options .InputFormatters.Add(new XmlDataContractSerializerInputFormatter(options ));
//more output formatters
var jsonOutputFormatter = options.OutputFormatters.OfType<JsonOutputFormatter>().FirstOrDefault();
if (jsonOutputFormatter != null)
{
jsonOutputFormatter.SupportedMediaTypes.Add("application/vnd.myvendormediatype");
}
}
}
But going back on the content negotiation in your controllers you can keep just one. The only thing is that you need to know the mediaType to return your View or your json content. Only be sure to pass an accept header with the content type you want. With the content type you are defining for an api or for an mvc application which is the content/format the client should expect
[HttpGet("[action]/{id}")]
public IActionResult public Product GetById(int id, [FromHeader(Name = "Accept")] string mediaType)
{
if (mediaType == "application/vnd.myvendormediatype")
{
var data = GetYourData(...)
return Json(data);
}
else return View("YourDefaultView");
}

What happening behind scene when we want to render view?

ActionResult is the base class for the various return types to View in MVC. So your action must return an ActionResult or a class derived from it in order to work.
so we can use
public ContentResult Index()
{
return Content("Hello world");
}
or for example
public ViewResult Index()
{
return View();
}
or ActionResult
public ActionResult Index()
{
if (ViewBag.Hello = "World")
return Json("");
return PartialView();
}
BUT also is possible use string !!!
public string Index()
{
return "Hello World";
}
WHY is than not possible return integer to view? (Or maybe it is?)
public int Index()
{
return 4;
}
and not possible return some entity to view (Or maybe it is?)
public User Index()
{
return new User();
}
My question is : What happening behind scene when we want to render view?

I agree that this is quite a broad question, but I wanted address and answer a few of the points you raised in your question.
You can return an int, string or object from your action method and it will simply return the object's string representation as the result.
Therefore you don't have to return an object of type ActionResult in order for it to work, but the ActionResult enables useful functionality through it's various implementations so that ASP.NET MVC Framework can handle different scenarios straight out of the box.
Such as returning views and handling the ViewModel you want to pass to your view:
return View(); // Default view without view model
return View(viewModelObject); // Default view with a view model
Returning views based on your routing information:
return View("viewName", viewModelObject);
Performing redirects to another page, using your routing information:
return RedirectToAction(actionName, controllerName);
Returning a page with specific status codes:
return new HttpStatusCodeResult(HttpStatusCode.BadRequest);
Returning JSON instead of a view:
return JsonResult(myObject);
All of the above examples do different things, return different types of results and handle your objects for you so that you don't have to code the behaviour yourself - they're ready for you to use.
Another handy thing is that you can create your own implementations of ActionResult to create your own behaviour, so it's very flexible in that regard.
I agree with #Daniel J.G. that you should do some more reading on how ASP.NET MVC hangs together and it will become a lot more clear to you.

ASP.NET MVC Return Overloaded Get Action

I am using MVC 4, and I have the following:
[HttpGet]
public ActionResult SomeForm(modelType model = null)
{
if(model != null)
return View(model);
return View(getModelFromSomewhere());
}
[HttpPost]
public ActionResult SomeForm(modelType model)
{
if(isValid())
doSomething();
else
return SomeForm(model) // Line in Question
}
However, obviously, I am getting an ambiguous method error on "Line in Question". I'm wondering if anyone has an elegant solution to be able to specify to return specifically the [Get] method of the same name?
Thank you!

You can't have methods with the same signature as you've pointed out already. In C# it also means you can't distinguish functions by just return type - so you must use different names if parameters are same (again default values are ignored when matching of signatures).
If you want separate GET and POST handler - use different names of methods and ActionNameAttribute to name the action:
[HttpGet]
[AciontName("SomeForm")]
public ActionResult SomeFormGet(modelType model = null) ...
[HttpPost]
[AciontName("SomeForm")]
public ActionResult SomeFormPost(modelType model) ...

make it compile...
[HttpPost]
public ActionResult SomeForm(modelType model, FormCollection fc)
{
if(isValid())
doSomething();
else
return SomeForm(model) // Line in Question
}

If you are using http get method you are waiting that browser will send you serialized model as a string query. For example, you are waiting url like
http://example.com?name=Andrew&type=Worker&field1=param1&field2=param2&....
It is common practice to use only id in your get method, so you can do it like this:
[HttpGet]
public ActionResult SomeForm(int id)
{
var model = FindModelById(id);
if(model != null)
return View(model);
return View(getModelFromSomewhere());
}
If you are looking for an elegant solution, it will be more elegant in architecture

Overloading Controller methods with custom JSON(POST) binding in MVC3

I reached a road bump while trying to use some of the cool new feature of MVC 3.
Is it possible to overload controllers using custom JSON binding using MVC3 ?
It doesnt look like it works automatically as of now....
What's the neatest way to do this ?
For example
If I want to implement the following endpoint
[HttpPost]
public ActionResult GetPet(Cat catObject)
{
return Json(catObject.purr());
}
overloaded with this endpoint
[HttpPost]
public ActionResult GetPet(Dog dogObject)
{
return Json(dog.bark());
}
Is there any way i can do this without using thirdparty libraries Or System.Web.Script.Serialization.JavaScriptSerializer
Also Is there any particular reason this is not implemented in mvc3 yet?

Overloading Json objects on controllers seems a distant possibility now rather than a present reality.
This is the closest I could get to overloading the action
[HttpPost]
public ActionResult GetPet()
{
Cat catObj;
Dog dogObg;
if (TryUpdateModel(catObj))
return Json(catObj.purr());
else
{
ModelState.Clear();
if (TryUpdateModel(dogObg))
return Json(dogObj.bark());
else
{
ModelState.Clear();
ModelState.AddModelError("InvalidInput", "The given input does not match with any of the accepted JSON types");
return new HttpBadRequestResult(ModelState);
}
}
}

ASP.NET MVC: How to create an action filter to output JSON?

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.

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();
}

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