假设您有两个从抽象基类派生的实体,并且您想要实现每个具体类型的表。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; }
}
并且您不想在抽象基类(EntityBase)中使用属性,您只想在dbcontext中为所有实体映射EntityBase类一次。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);
});
}
}
以下是您的问题的答案。。
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,您可以为继承自EntityBase的每个表编写具体的配置类,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());
}
}