使用EF Core包含所有继承的类的所有属性

Question

When using EF6, I was using lazy loading so I never had this issue, but with EFCore, I don't know if this is possible with a single query.

I have the following class structure

class A { public B b; }
class B { public ICollection<C> list_c; }
class C { public ICollection<D> list_d; }
abstract class D { public long c_id; }
class Da { public E e; }
class Db { public F f; }

I need a list of all the D objects, but with access to their e and f properties respectively. I have a working query at the moment where I query _db.D over a list of c_id 's that I fetch using the first half of the query below, but with that approach, I send one query to get all the c_id 's and then one query per type (I have 4 types).

I was wondering if I can make it work with one call that looks something like this:

_db.As.Include(x => x.b)
      .ThenInclude(x => x.list_c)
      .ThenInclude(x => x.list_d)
      // some magic here
      .FirstOrDefaultAsync(x=> x.Id = model.Id);

EDIT:

At the moment this is how I make the list:

var a = await _db.As.Include(x => x.b)
                    .ThenInclude(x => x.list_c)
                    .FirstOrDefaultAsync(x=> x.Id = model.Id);

var result = await _db.Ds.OfType<Da>()
                         .Include(x=>x.e)
                         .Where(x=>a.b.list_c.Any(y=>y.Id == x.c_id))
                         .Select(x=>(D)x)
                   .Concat(_db.Ds.OfType<Db>()
                         .Include(x=>x.f)
                         .Where(x=>a.b.list_c.Any(y=>y.Id == x.c_id))
                         .Select(x=>(D)x)).
                   .ToListAsync();

Answer 1

I would advise against using Entity Framework Core if you have that option. Check the roadmap at https://github.com/aspnet/EntityFrameworkCore/wiki/Roadmap - It currently doesn't have lazy loading to name of its (very annoying) shortcomings.

You can have your data access layer in a .NET 4.6 project and have your consuming project(s), even if the consuming is based on Core, reference that project without difficulty.