ASP.NET MVC 3 - Create a dynamic object only for return

Question

I have a javascript application (ASP.net MVC 3 on the back end with SignalR) and i want to return only the fields that are needed (it must be dynamic).

I use the entity framework but i cant select only the specific columns because i need some fields for checking or something in the method that are no needed on the client side.

so, currently i make this:

    public void GetPerson(int personID)
    {
        // Some logic...

        // person is a entity from the entity framework (Person)
        dynamic p = new
        {
            ID = person.ID,
            FirstName = person.FirstName,
            LastName = person.LastName
        };

        Clients[Context.ConnectionId].loadPerson(p);
    }

But i'm not sure, if its a good practice (Performance etc.). Is there a better solution or can i still continue with the dynamic type?

I don't understand your example. You create p, but don't use it? — Erik Funkenbusch, Mar 02 '12 at 20:42
This would he fine (assuming you return p at some point). I just wonder if the bandwidth you save is really all that much. If your are returning many objects, or you have high traffic it is probably worth it, but if neither of these are the case then it might not be necessary. — Mike Marshall, Mar 02 '12 at 20:49
@mjmarsh, yes, it can be true that i must return 30'000 dynamic objects — Wasser Trinker, Mar 02 '12 at 20:52

score 2 · Accepted Answer · edited May 23 '17 at 12:29

Are you using the ExpandoObject?

Going by your example, maybe use just an anonymous type instead of a dynamic type.

 var p = new
        {
            ID = person.ID,
            FirstName = person.FirstName,
            LastName = person.LastName
        };

This will make sure that you still get compile-time checking. However, either should work. See this for more detail on anonymous vs dynamic.

score 0 · Answer 2 · answered Mar 02 '12 at 20:44

0

Looks fine to me. If you need the other columns then return them and do what you need to do with them and then as you are doing create a anonymous person and pass it in.

answered Mar 02 '12 at 20:44

Jon

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score 0 · Answer 3 · answered Mar 02 '12 at 21:04

The best practice with MVC, whether returning data to a view or to the client via JSON or even using SignalR, is to use a strongly typed object. Just create a ViewModel specifically for the data you want to return and use that. I usually name these classes using a convention of [Controller][Action]ViewModel. So one might be MemberDetailsViewModel.

The benefits include the compile time checking, easier serialization and, less problems when it comes to modifications later.

I tried exactly what your doing when I first started playing around with MVC. trust me, it's not worth the convenience you perceive to be getting now.

If your data being returned MUST be dynamic, you may want to try returning a collection of key value pairs.

ASP.NET MVC 3 - Create a dynamic object only for return

3 Answers3