Question

I have a table with 9 columns in database and I want to be able to load only some fields of it if I need.

How can I do this with Entity Framework 4 please?

e.g. My table has these fields:

ID, FirstName, LastName, FotherName, BirthDate, Mobile, Email

and I want to be able to fetch just these columns:

ID, FirstName, LastName

My project is an ASP.NET MVC 3 application, with SQLServer 2008 Express and EF 4.1.

Answer 1

Assume you have a table with this model:

public class User{
    public int ID {get; set;}
    public string NickName {get; set;}
    public string FirstName {get; set;}
    public string LastName {get; set;}
    public string FotherName {get; set;}
    public DateTime BirthDate {get; set;}
    public string Mobile {get; set;}
    public string Email {get; set;}
    public string Password {get; set;}
}

Now, you want fetch just ID, FirstName, LastName, and FotherName. You can do it in 2 way; The first way is fetch them as an anonymous object, look:

var user = entityContext.Users.Where(u => u.ID == id)
    .Select(u => new {
        ID = u.ID,
        FirstName = u.FirstName,
        LastName = u.LastName,
        FotherName = u.FotherName
    }).Single();

Now, your return-value-type is anonymous, you can work with it such as:

var i = user.ID;
// or
var s = user.FirstName;

In another way (for example when you want to pass the object as an Model to a View), you can define a new class, (i.e. UserViewModel), and when you select the object, select it as a UserViewModel. look:

public class UserViewModel{
    public int ID {get; set;}
    public string NickName {get; set;}
    public string FirstName {get; set;}
    public string LastName {get; set;}
    public string FotherName {get; set;}
}

and in query, take this:

var user = entityContext.Users.Where(u => u.ID == id)
    .Select(u => new UserViewModel {
        ID = u.ID,
        FirstName = u.FirstName,
        LastName = u.LastName,
        FotherName = u.FotherName
    }).Single();

Look that just ONE difference is between them, in labda expression, instead of u => new {}
we are using u => new UserViewModel{}. Good luck.