EF Core实现了Table-Per-Concrete-Type，具有抽象基类的流畅映射

Question

假设您有两个派生自抽象基类的实体，并且您希望实现Table-Per-Concrete-Type。 以下实体：

public abstract class EntityBase
{
   public int Id { get; set; }

   public string CreatedBy { get; set; }

   public DateTime CreatedAt { get; set; }
}

public class Person : EntityBase
{
   public string Name { get; set; }
}
public class PersonStatus : EntityBase
{
   public string Title { get; set; }
}

并且您不希望在抽象基类（EntityBase）中使用属性，您只想在dbcontext中为所有实体映射一次 EntityBase类。 如何更改以下代码：

public class PeopleDbContext : DbContext
{
   public DbSet<Person> People { get; set; }

   protected override void OnModelCreating(ModelBuilder modelBuilder)
   {
      base.OnModelCreating(modelBuilder);

      // Entity base class mapping(only once)

      modelBuilder.Entity<Person>(e =>
      {
        e.Property(x => x.Name)
            .IsRequired()
            .HasMaxLength(100);
      });
      modelBuilder.Entity<PersonStatus>(e =>
      {
        e.Property(x => x.Title)
            .IsRequired()
            .HasMaxLength(100);
      });
  }

}

Answer 1

这是你的问题的答案..

您需要为BaseClass编写配置：

public class EntityBaseConfiguration<TBase> : IEntityTypeConfiguration<TBase>
    where TBase : EntityBase
{
    public virtual void Configure(EntityTypeBuilder<TBase> builder)
    {
        builder.HasKey(b => b.Id);
        builder.Property(b => b.CreatedBy)
            .HasColumnType("varchar(50)");
        builder.Property(b => b.CreatedAt)
            .HasColumnType("datetime2");
    }
}

之后，您可以编写具体的Configuration-Class foreach Table，它继承自EntityBase，如下所示：

public class PersonConfig : BaseConfig<Person>
{
    public override void Configure(EntityTypeBuilder<Person> builder)
    {
        base.Configure(builder);
        builder.Property(e => e.Name)
            .HasColumnType("varchar(100)")
            .IsRequired();
    }
}

要在dbContext中调用配置，可以调用ApplyConfiguration：

public class PeopleDbContext : DbContext
{
    public DbSet<Person> People { get; set; }

    protected override void OnModelCreating(ModelBuilder modelBuilder)
    {
        modelBuilder.ApplyConfiguration(new PersonConfig());
    }
}