基本上我有一个像这样的实体:

public class Person {
 public int PersonId { get; set; }
 public string Name { get; set; }
 public Address Hometown { get; set; }
}

还有一个类:

public class Address {
 public City City { get; set; }
 public string Province { get; set; }
}

我想要完成的是垂直连接两个类并有一个包含行的表:

TB_PERSON:
   PersonId PK
   Name
   City_id FK
   Province

我想要这种方法的原因是，在我的实际项目中，我在多个条目上出现了相同类型的数据结构模式，在这种情况下，示例就是地址类。它可能很容易出现在另一个实体中。

我好几天都找不到如何做到这一点是不是太难了？我能得到的最接近的是复杂类型，但在这种情况下它们不允许导航属性。我想访问并拥有我的行数据类型的结构化和面向对象，我认为 EF 可以尝试。感谢您的帮助。

最佳答案

ComplexType 应该是一个解决方案，但不幸的是:

复杂类型不能包含导航属性。 Source

解决方法列表:

表拆分的解决方法

public class Person
{
    public int PersonID { get; set; }
    public string Name { get; set; }
    public virtual Address Address { get; set; }
}

public class Address
{
    public Int32 ID { get; set; }
    public string Province { get; set; }
    public virtual City City { get; set; }

}

public class City
{
    public Int32 CityID { get; set; }
    public string Name { get; set; }
}

public class MappingContext : DbContext
{
    public DbSet<Person> Persons { get; set; }

    public DbSet<Address> Addresses { get; set; }

    protected override void OnModelCreating(DbModelBuilder modelBuilder)
    {
        modelBuilder.Entity<Address>()
            .HasKey(t => t.ID)
            .HasOptional(t => t.City)
            .WithMany()
            .Map(t => t.MapKey("CityID"));

        modelBuilder.Entity<Address>()
            .Property(t => t.ID)
            .HasColumnName("PersonID");

        modelBuilder.Entity<Person>()
            .HasKey(t => t.PersonID)
            .HasRequired(t => t.Address)
            .WithRequiredPrincipal();

        modelBuilder.Entity<Person>().ToTable("TB_PERSON");

        modelBuilder.Entity<Address>().ToTable("TB_PERSON");

        modelBuilder.Entity<City>()
            .HasKey(t => t.CityID)
            .ToTable("City");
    }
}

[用法]

    using (var db = new MappingContext())
    {
        var person = db.Persons.FirstOrDefault();
        var cityName = person.Address.City.Name;

        var address = db.Addresses.FirstOrDefault();
        var personName = address.Person.Name;
    }

[数据库]

    CREATE TABLE [dbo].[City](
        [CityID] [int] IDENTITY(1,1) NOT NULL,
        [Name] [varchar](50) NULL
    ) ON [PRIMARY]

    CREATE TABLE [dbo].[TB_PERSON](
        [PersonId] [int] IDENTITY(1,1) NOT NULL,
        [Name] [varchar](50) NULL,
        [Province] [varchar](50) NULL,
        [CityID] [int] NULL
    ) ON [PRIMARY]

Table Splitting + TPC 继承的解决方法(对于可重用的 Address 类)

TB_CUSTOMER 是另一个包含地址列的表。

public class Person
{
    public int PersonID { get; set; }
    public string Name { get; set; }
    public virtual PersonAddress Address { get; set; }
}

public class Address
{
    public string Province { get; set; }
    public virtual City City { get; set; }

}

public class PersonAddress : Address
{
    public Int32 PersonID { get; set; }
    public virtual Person Person { get; set; }
}
public class CustomerAddress : Address
{
    public Int32 CustomerID { get; set; }
}

public class Customer
{
    public int CustomerID { get; set; }
    public string Name { get; set; }
    public virtual CustomerAddress Address { get; set; }
}

public class City
{
    public Int32 CityID { get; set; }
    public string Name { get; set; }
}

public class MappingContext : DbContext
{
    public DbSet<Person> Persons { get; set; }
    public DbSet<Customer> Customers { get; set; }
    public DbSet<PersonAddress> PersonAddresses { get; set; }
    public DbSet<CustomerAddress> CustomerAddresses { get; set; }

    protected override void OnModelCreating(DbModelBuilder modelBuilder)
    {
        modelBuilder.Entity<PersonAddress>()
            .HasKey(t => t.PersonID)
            .HasOptional(t => t.City)
            .WithMany()
            .Map(t => t.MapKey("CityID"));

        modelBuilder.Entity<CustomerAddress>()
            .HasKey(t => t.CustomerID)
            .HasOptional(t => t.City)
            .WithMany()
            .Map(t => t.MapKey("CityID"));

        modelBuilder.Entity<Person>()
            .HasRequired(t => t.Address)
            .WithRequiredPrincipal(t => t.Person);

        modelBuilder.Entity<Customer>()
            .HasRequired(t => t.Address)
            .WithRequiredPrincipal();

        modelBuilder.Entity<Person>().ToTable("TB_PERSON");
        modelBuilder.Entity<PersonAddress>().ToTable("TB_PERSON");

        modelBuilder.Entity<Customer>().ToTable("TB_CUSTOMER");
        modelBuilder.Entity<CustomerAddress>().ToTable("TB_CUSTOMER");

        modelBuilder.Entity<City>()
            .HasKey(t => t.CityID)
            .ToTable("City");
    }
}

使用 IAddress 的解决方法

public class Person : IAddress
{
    public int PersonID { get; set; }
    public string Name { get; set; }
    public string Province { get; set; }
    public virtual City City { get; set; }

    [NotMapped]
    public IAddress Address { get { return this; } }
}

public interface IAddress
{
    string Province { get; set; }
    City City { get; set; }

}

public class City
{
    public Int32 CityID { get; set; }
    public string Name { get; set; }
}

public class MappingContext : DbContext
{
    public DbSet<Person> Persons { get; set; }

    protected override void OnModelCreating(DbModelBuilder modelBuilder)
    {

        modelBuilder.Entity<Person>()
            .HasKey(t => t.PersonID)
            .HasOptional(t => t.City)
            .WithMany()
            .Map(t => t.MapKey("CityID"));

        modelBuilder.Entity<Person>().ToTable("TB_PERSON");

        modelBuilder.Entity<City>()
            .HasKey(t => t.CityID)
            .ToTable("City");
    }
}

关于c# - 我如何将类属性(带有导航 Prop )作为实体属性？复杂类型不行，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/42403179/