ASP.NET MVC Passing Two Models to Details View

Question

This question has probably been answered a million times, however, I spent over three hours and I can't find an answer to my problem. I am trying to pass two models to my Details view and I have trouble understanding what shall be returned by my Details controller.

These are my models:

public class Property
{
    public int PropertyID { get; set; }
    public string Address { get; set; }
    public string City { get; set; }
    public string ProvinceState { get; set; }
    public string ZipCode { get; set; } 
    public string Country { get; set; }

}

public class PropertySimilar {
    public IEnumerable<Property> Properties { get; set; }
    public Property CurrentProperty { get; set; }
}

This is my controller:

public ActionResult Details(int? id)
    {
        if (id == null)
        {
            return new HttpStatusCodeResult(HttpStatusCode.BadRequest);
        }

        Property property = db.Properties.Find(id);

        if (property == null)
        {
            return HttpNotFound();
        }
        return View(db.Properties.ToList());
    }

I am trying to display the select property, in addition to, three other random properties underneath it.

Any guidance would be greatly appreciated. Thanks.

The simple rule of thumb in such cases is to leverage the viewmodel concept.Keep all the displayed properties in a viewmodel and pass that viewmodel to view. — Navoneel Talukdar, Jan 23 '18 at 03:48
You can put both models in a single viewmodel class, and then pass required properties to view. Similar issues can be found [here](https://stackoverflow.com/questions/944334/asp-net-mvc-view-with-multiple-models) and [here](https://stackoverflow.com/questions/17030399/pass-two-models-to-view). — Tetsuya Yamamoto, Jan 23 '18 at 03:50
Thank you both for the speedy reply, I get it now! So simple.. — HereToLearn, Jan 23 '18 at 04:09

score 0 · Accepted Answer · edited Jan 23 '18 at 04:12

With Neel's and Tetsuya's help, I was able to solve my issue. Here is what the controller should look like:

public ActionResult Details(int? id)
{
    if (id == null)
    {
        return new HttpStatusCodeResult(HttpStatusCode.BadRequest);
    }

    var property = db.Properties.Find(id);

    if (property == null)
    {
        return HttpNotFound();
    }

    PropertySimilar pros = new PropertySimilar();
    pros.CurrentProperty = property;
    pros.Properties = db.Properties.ToList();

    return View(pros);
}

ASP.NET MVC Passing Two Models to Details View

1 Answers1