Is my view model too "complex" to allow binding on post?

Question

Using my model displaying a page works fine but the post does not return the bound model. My classes:

public class ContactManager
{       
    public Contact Contact { get; set; }       
    public SelectList SalutationList { get; set; }
}
public class Contact
{
    public int Id{get;set;}
    public string FirstName{get; set;}
    public SalutationType SalutationType{get; set;}
}
 public class SalutationType
{      
    public int Id { get; set; }
    public string Name { get; set; }      
}

My View:

@model ViewModels.ContactManager

@using (Html.BeginForm())
{
    @Html.AntiForgeryToken()
    @Html.ValidationSummary(true)
    @Html.HiddenFor(model => model.Contact.Id)
    @Html.DropDownListFor(model => model.Contact.SalutationType.Id, Model.SalutationList)
    @Html.EditorFor(model => model.Contact.FirstName)
    <input type="submit" value="Save" />
}

The issue seems to be in the DropDownListFor. The dropdown list displays correctly with the proper value but when I post this page the complete model is blank. If I simplify the DropDownListFor like this the values are posted as expected.

@html.DroDownListFor(model=>model.MyPlaceHolderProp, Model.SalutationList)

Is my model too complex? Am I not doing something correctly?

The models are based off of several tables using EF that I have created in a separate project. I am trying to avoid creating more classes/models then I have to.

Answer 1

I am not sure this will help you... I wsa having a similar issue but I was using ajax to post back... anyway, I had forgotten to mark my binding class with the [Serializable] attribute.

so you might try

[Serializable]
public class Contract {
  ...
}

Again, I am using Json to post back to my controller so may not be related or help you. But, I guess could be worth a try.

Answer 2

Have you considered using a custom model binder? Custom model binding isn't all that complicated for models that are still relatively simple, and you can handle the serialization/deserialization however you need to.

http://msdn.microsoft.com/en-us/magazine/hh781022.aspx

http://ivonna.biz/blog/2012/2/2/custom-aspnet-model-binders-series,-part-3-subclassing-your-models.aspx

http://forums.asp.net/t/1944696.aspx?what+is+custom+model+binding+in+mvc

Answer 3

You should post your controller action as well, as your model coming back as blank really has nothing to do with this. Changing the DropDownListFor definition one way or another should not effect the posting of any other values.

That said, you will run into another issue eventually here, so you need to regroup, anyways. You can't just post back the id value of a related item. Entity Framework will either complain that there's already an object with that id, or worse, if the object attaches, it will update the row with that id with the new posted value for Name, which in this case, is nothing, so it'll just clear it out.

When you create a relationship with a single item (a foreign key basically), if you don't specify a property to hold that foreign key value, Entity Framework creates one for you behind the scenes to track the relationship. In your case here, that means your Contacts table has a column named SalutationType_Id. However, there's no way from your class to directly access this value. This is why I recommend that you always provide an explicit property to handle the relationship:

[ForeignKey("SalutationType")]
public int SalutationTypeId { get; set; }

public SalutationType SalutationType { get; set; }

If you do that, then you can directly stuff the posted id there and Entity Framework will create the relationship.

@Html.DropDownListFor(m => m.Contact.SalutationTypeId, Model.SalutationList);

If you insist on keeping the key implicit, then you must create the relationship yourself, by creating a field on your view model to hold the posted value, then using that value to look up the SalutationType instance from the database, and then finally adding that to the Contact instance.

Add to your view model

public int SalutationTypeId { get; set; }

In your view

@Html.DropDownListFor(m => m.SalutationTypeId, Model.SalutationList)

In your POST action

var salutationType = db.SalutationTypes.Find(model.SalutationTypeId);
contact.SalutationType = salutationType;

Answer 4

You could do it this way. This may be the more "MVC best practice" way to handle it. Everything stays neatly in their models, and no manual IDs are required. The views are intended to be representations of the underlying models they are built on. If you are creating a view that has a form, then create a model that represents the form and use it in the view.

Revise your models like:

public class PostModel
{
    public int ContactID { get; set; }
    public int SalutationID { get;  set; }
    public string FirstName { get; set; }
}

public class PostView
{
    public ContactManager contact { get; set; }
    public PostModel post { get; set; }
}

Then create the PostView in the controller:

public ActionResult Index() 
{
    //create the PostView model

    var pv = new PostView();
    pv.ContactManager = contactManager;
    pv.post = new PostView()
    {
        ContactID = contactManager.Contact.Id,
        SalutationID = contactManager.SalutationType.Id,
        FirstName = contactManager.Contact.FirstName
    };

    return View(pv);
}

Then the view could be like:

@model ViewModels.PostView

@using (Html.BeginForm())
{
    @Html.AntiForgeryToken()
    @Html.ValidationSummary(true)
    @Html.HiddenFor(model => model.post.ContactID)
    @Html.DropDownListFor(model => model.post.SalutationID, model.contact.SalutationList)
    @Html.EditorFor(model => model.post.FirstName)
    <input type="submit" value="Save" />
}

Then the post action in the controller:

[HttpPost]
public ActionResult Index(PostView pv) 
{
    //post code
    //the posted data will be in pv.post
}