Question

I have a simple 1:many aggregate relationship, lets say:

public class Parent
{
    public string Name {get; set;}
    public Child SelectedChild {get; set;}
    public Child PublishedChild {get; set;}
    public virtual ICollection<Child> AllChildren {get; set;}
}

public class Child
{
    public string Name {get; set;}
    [Required]
    public Parent Father {get; set;}
}

When creating the schema from this model I get the error:

Introducing FOREIGN KEY constraint ‘Parent_SelectedChild’ on table ‘Parent’ may cause cycles or multiple cascade paths.
Specify ON DELETE NO ACTION or ON UPDATE NO ACTION, or modify other FOREIGN KEY constraints

So I add the following to OnModelCreating:

        modelBuilder.Entity<Child>()
            .HasRequired(v => v.Parent)
            .WithOptional(c => c.SelectedChild)
            .WillCascadeOnDelete(false);

        modelBuilder.Entity<Child>()
            .HasRequired(v => v.Parent)
            .WithOptional(c => c.PublishedChild)
            .WillCascadeOnDelete(false);

This gets round the original error but I now get:

Unable to determine the principal end of the ‘xxx.Parent_SelectedChild’ relationship.
Multiple added entities may have the same primary key.

Can anyone help?
All I essentially want to do is refer to particular child records on a 1:many aggregate relationship from the parent. I assume EF will create INT child id columns on the parent called e.g. SelectedChild_Id & PublishedChild_Id (or similar).

Thanks in advance
-macon

Edit: In response to @Slauma:
I can get a schema generated using:

            modelBuilder.Entity<Parent>()
            .HasOptional(p => p.SelectedChild)
            .WithOptionalPrincipal()
            .WillCascadeOnDelete(false);

        modelBuilder.Entity<Parent>()
            .HasOptional(p => p.PublishedChild)
            .WithOptionalPrincipal()
            .WillCascadeOnDelete(false);

        modelBuilder.Entity<Parent>()
            .HasMany(p => p.AllChildren)
            .WithRequired(c => c.Father)
            .WillCascadeOnDelete(false);

But this generates multiple FK on the Child record e.g. Parent_Id, Parent_Id1. I just want a reference from the Parent to one of the child rows e.g. Parent_SelectedChildId. Do I have to do this manually with an int column on parent?

Answer 1

I think you have three 1-to-many relationships:

modelBuilder.Entity<Parent>()
    .HasOptional(p => p.SelectedChild)
    .WithMany()
    .WillCascadeOnDelete(false);

modelBuilder.Entity<Parent>()
    .HasOptional(p => p.PublishedChild)
    .WithMany()
    .WillCascadeOnDelete(false);

modelBuilder.Entity<Parent>()
    .HasMany(p => p.AllChildren)
    .WithRequired(c => c.Father)
    .WillCascadeOnDelete(false);

Edit

I’ve tested my mapping above with exactly the Parent and Child class you provided in your question – with the only exception that I have added a primary key property to both classes: public int Id { get; set; }. Otherwise EF would complain about a missing key property. This mapping doesn’t throw an exception and creates the following tables in the database:

Parents table:

- Id                    int             not nullable (PK)
- Name                  nvarchar(MAX)   nullable
- SelectedChild_Id      int             nullable (FK)
- PublishedChild_Id     int             nullable (FK)

Children table:

- Id                    int             not nullable (PK)
- Name                  nvarchar(MAX)   nullable
- Father_Id             int             not nullable (FK)

So, there are the three foreign key columns as expected.

Since you get an exception according to your comment, I guess that there is actually some important difference in the code you have tested.

BTW: Mapping the two navigation properties of the Parent class as One-to-One relationships is much more difficult, if not impossible. In EF you need a shared primary key between the two tables to map a One-to-One relationship, so it would not be possible to assign two different entities to the two navigation properties because they cannot both have the same key as the parent.