Set the model depending on what it is expecting in view

Question

The goal

Change the model depending on what it is expecting.

The problem

I have a method in my ProductsControllercalled Category. When someone request this method, two things can happen: if the parameter passed to the method is different than Daily-Offers, then one type of list is passed to View. If the parameter passed to the method is equal to Daily-Offers, then another type of list is passed to View.

To better comprehension, see:

[HttpGet]
public ActionResult Category(string categoryName = null)
{
    int? categoryId = categoryName != "Daily-Offers" ? 
                      Convert.ToInt32(Regex.Match(categoryName, @"\d+").Value) :
                      (int?)null;

    if (categoryName == "Daily-Offers")
    {
        var productsList = Products.BuildOffersList();
        ViewBag.Title = String.Format("Today's deal: ({0})", DateTime.Now);
        ViewBag.CategoryProductsQuantity = productsList.Count;
        ViewBag.CurrentCategory = "Daily-Offers";
        return View(productsList);
    }
    else if (Regex.Match(categoryName, @"\d+").Success && 
             String.Format("{0}-{1}", 
             categoryId, 
             CommodityHelpers.UppercaseFirst
                (CommodityHelpers.GenerateSlug
                     (Categories.GetDetails((sbyte)categoryId).Category_Name))) 
            == categoryName)
    {
        ViewBag.Title = Categories.GetDetails((sbyte)categoryId).Category_Name;
        ViewBag.CategoryProductsQuantity = 
        Categories.GetDetails((sbyte)categoryId).Category_Products_Quantity;
        ViewBag.CurrentCategory = 
           CommodityHelpers.UppercaseFirst(CommodityHelpers.GenerateSlug
           (Categories.GetDetails((sbyte)categoryId).Category_Name));
        return View(Products.BuildListForHome(categoryId, null));
    }
    else
    {
        return View("404");
    }
}

As you can see, the view should be prepared to receive IList<Offers> or/and IList<NormalProducts> — and I do not know how to do this.

What I have already tried

Something like this:

@model if(IEnumerable<BluMercados.Models.Data.getProductsListForHome_Result>) 
       ?: (IEnumerable<BluMercados.Models.Data.getProductsInOfferList_Result>);

But, of course, no success. Just to illustrate.

Answer 1

A view should really only have one model, so trying to make a view consume two different models, while doable, shouldn't be done.

Instead, you can create different views. Say you create a new view for DailyOffers, which takes IEnumerable<Offers> as its model.

You can then user an overload of View() to specify which view to return:

return View("DailyOffers", productsList);

However, instead of doing this, would it make more sense to redirect to a different action in the case of DailyOffers? So, you have a new action:

public ActionResult DailyOffers(...)

and instead of return View(productsList) you do:

return RedirectToAction("DailyOffers");

This all assumes that the models are sufficiently different from one another. If they are similar, using the Interface solution as suggested by p.s.w.g. would make more sense.