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我有一个关于 EF 5.0 和继承对象映射到与超类相同的表的问题。

我的项目如下所示：

public class ContextA : DbContext
{
    public DbSet<EntityA> EntitiesA { get; set; }
    public DbSet<EntityB> EntitiesB { get; set; }

    public ContextA(string connnectionString) : base(connnectionString) { }

    protected override void OnModelCreating(DbModelBuilder modelBuilder)
    {
        ModelCreating(modelBuilder);
    }

    public static void ModelCreating(DbModelBuilder modelBuilder)
    {
        modelBuilder.Entity<EntityA>().ToTable("EntityAs");
        modelBuilder.Entity<EntityB>().ToTable("EntityBs");

        modelBuilder.Entity<EntityA>().HasKey(e => e.EntityAId);
        modelBuilder.Entity<EntityB>().HasKey(e => e.EntityBId);
    }
}

public class EntityA
{
    public int EntityAId { get; set; }
    public string Name { get; set; }
}

public class EntityB
{
    public int EntityBId { get; set; }
    public string Name { get; set; }

    public int EntityAId { get; set; }
}

很好很好。请记住，实体 A 和 B 可能位于数据库的不同模式中，或者可能位于不同的上下文中，这就是 statc ModelCreation 方法的原因。

现在我想组合这样的上下文：

public class ContextB : DbContext
{
    public DbSet<EntityAA> EntityAAs { get; set; }
    public DbSet<EntityBB> EntityBBs { get; set; }

    public ContextB(string connnectionString) : base(connnectionString) { }

    protected override void OnModelCreating(DbModelBuilder modelBuilder)
    {
        ModelCreating(modelBuilder);
    }

    public static void ModelCreating(DbModelBuilder modelBuilder)
    {
        ContextA.ModelCreating(modelBuilder);

        modelBuilder.Entity<EntityAA>().Map(m =>
        {
            m.MapInheritedProperties();
            m.ToTable("EntityAs");
        }).HasKey(e => e.EntityAId)
        .Property(e => e.EntityAId)
        .HasDatabaseGeneratedOption(DatabaseGeneratedOption.Identity);

        modelBuilder.Entity<EntityBB>().Map(m =>
        {
            m.MapInheritedProperties();
            m.ToTable("EntityBs");
        })
        .HasKey(e => e.EntityBId)
        .Property(e => e.EntityBId)
        .HasDatabaseGeneratedOption(DatabaseGeneratedOption.Identity);

        modelBuilder.Entity<EntityAA>()
                    .HasMany(ea => ea.EntityBs)
                    .WithRequired(eb => eb.EntityA)
                    .HasForeignKey(eb => eb.EntityAId);
    }
}

public class EntityAA : EntityA
{
    private readonly List<EntityBB> _entityBbs = new List<EntityBB>();

    public virtual ICollection<EntityBB> EntityBs
    {
        get { return _entityBbs; }
    }
}

public class EntityBB : EntityB
{
    public EntityAA EntityA { get; set; }
}

我想这样做，因为我想拥有特定于模式的上下文，我想将它们组合到特定于应用程序的上下文中。

现在的问题是，我在模型创建过程中遇到错误：

Ergebnis Meldung:   Die EfInheritance.Test.GlobalTests.InsertEntitiesContextB-Testmethode hat eine Ausnahme ausgelöst: System.InvalidOperationException: The foreign key component 'EntityAId' is not a declared property on type 'EntityBB'. Verify that it has not been explicitly excluded from the model and that it is a valid primitive property.
Ergebnis StackTrace:    
bei System.Data.Entity.ModelConfiguration.Configuration.Properties.Navigation.ForeignKeyConstraintConfiguration.Configure(EdmAssociationType associationType, EdmAssociationEnd dependentEnd, EntityTypeConfiguration entityTypeConfiguration)
   bei System.Data.Entity.ModelConfiguration.Configuration.Properties.Navigation.NavigationPropertyConfiguration.ConfigureConstraint(EdmAssociationType associationType, EdmAssociationEnd dependentEnd, EntityTypeConfiguration entityTypeConfiguration)
   bei System.Data.Entity.ModelConfiguration.Configuration.Properties.Navigation.NavigationPropertyConfiguration.ConfigureDependentBehavior(EdmAssociationType associationType, EdmModel model, EntityTypeConfiguration entityTypeConfiguration)
   bei System.Data.Entity.ModelConfiguration.Configuration.Properties.Navigation.NavigationPropertyConfiguration.Configure(EdmNavigationProperty navigationProperty, EdmModel model, EntityTypeConfiguration entityTypeConfiguration)
   bei System.Data.Entity.ModelConfiguration.Configuration.Types.EntityTypeConfiguration.ConfigureAssociations(EdmEntityType entityType, EdmModel model)
   bei System.Data.Entity.ModelConfiguration.Configuration.Types.EntityTypeConfiguration.Configure(EdmEntityType entityType, EdmModel model)
   bei System.Data.Entity.ModelConfiguration.Configuration.ModelConfiguration.ConfigureEntities(EdmModel model)
   bei System.Data.Entity.DbModelBuilder.Build(DbProviderManifest providerManifest, DbProviderInfo providerInfo)
   bei System.Data.Entity.DbModelBuilder.Build(DbConnection providerConnection)
   bei System.Data.Entity.Internal.LazyInternalContext.CreateModel(LazyInternalContext internalContext)
   bei System.Data.Entity.Internal.RetryLazy`2.GetValue(TInput input)
   bei System.Data.Entity.Internal.LazyInternalContext.InitializeContext()
   bei System.Data.Entity.Internal.InternalContext.GetEntitySetAndBaseTypeForType(Type entityType)
   bei System.Data.Entity.Internal.Linq.InternalSet`1.Initialize()
   bei System.Data.Entity.Internal.Linq.InternalSet`1.get_InternalContext()
   bei System.Data.Entity.Internal.Linq.InternalSet`1.ActOnSet(Action action, EntityState newState, Object entity, String methodName)
   bei System.Data.Entity.Internal.Linq.InternalSet`1.Add(Object entity)
   bei System.Data.Entity.DbSet`1.Add(TEntity entity)
   bei EfInheritance.Test.GlobalTests.InsertEntitiesContextB() in c:\temp\EfInheritance\EfInheritance.Test\GlobalTests.cs:Zeile 45.

最后，我的问题是：

  1. 有可能像这样工作吗？
  2. 如果这是一个可能的解决方案，我该怎么做才能让它工作？

如果我在 ModelCreation 中注释 ForeignKey-Statement，它可以正常工作，除了运行简单测试后数据库中的结果：

using (var ctx = new ContextB("CodeFirstDatabase"))
        {
            EntityAA aa = new EntityAA();
            aa.Name = "Test_AA_01";

            EntityBB bb = new EntityBB();
            bb.Name = "Test_BB_01";

            aa.EntityBs.Add(bb);

            ctx.EntityAAs.Add(aa);

            ctx.SaveChanges();
        }

EntityBs EntityBId | 姓名 | 实体 ID | 鉴别器 | EntityA_EntityAId 1 | 测试_BB_01 | 0 | 实体BB | 1

EntityAs EntityAId | 姓名 | 鉴别器 1 | 测试_AA_01 | 实体AA

现在还有一个我无法访问的额外 ForeignKeyColumn 以及一个额外的鉴别器列。

有人知道该怎么做才能使其正常工作吗？如有必要，我已准备好将项目邮寄/附加。

亲切的问候，

斯文

4

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