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编辑：看起来这在实体框架中被称为“复杂类型”，我看到我如何能够在列上设置名称以在类中包含一次复杂类型，但我不知道如何/如果你可以使用类中包含的多个相同的复杂类型来做到这一点。

http://blogs.msdn.com/b/wriju/archive/2010/07/26/ef4-complex-type.aspx

EntityFramework 4.1 Code First 错误地命名复杂类型列名

由于 Castle ActiveRecord 已经日落，我正在转向 Entity Framework 5，但是我不知道如何实现 Activerecord 的“嵌套属性”概念：

http://docs.castleproject.org/Active%20Record.Mappings.ashx#Nested_5

http://docs.castleproject.org/Default.aspx?Page=Nested%20data%20%28NHibernate%20components%29&NS=Active%20Record&AspxAutoDetectCookieSupport=1

我已经完成了谷歌搜索，但我不确定“嵌套”是否是 EF 中这个概念的正确术语，如果它甚至存在的话。这是我在 Castle ActiveRecord 中的表格和课程，有人可以为此提供 EF 术语以及如何完成它：

AmaStatementEntity 上的 Address 属性标记为<Nested(ColumnPrefix:="FL")))>映射到数据库列 FLStreet、FLSuite、FLCity 等。

CREATE TABLE [dbo].[AMA](
--other columns not necessary to discussion removed
    [AMAId] [uniqueidentifier] NOT NULL,
    [FLStreet] [nvarchar](255) NULL,
    [FLSuite] [nvarchar](255) NULL,
    [FLCity] [nvarchar](255) NULL,
    [FLZipCode] [nvarchar](255) NULL,
    [FLCountry] [nvarchar](255) NULL,
    [FLState] [nvarchar](255) NULL,
    [BAStreet] [nvarchar](255) NULL,
    [BASuite] [nvarchar](255) NULL,
    [BACity] [nvarchar](255) NULL,
    [BAZipCode] [nvarchar](255) NULL,
    [BACountry] [nvarchar](255) NULL,
    [BAState] [nvarchar](255) NULL,
    [PAStreet] [nvarchar](255) NULL,
    [PASuite] [nvarchar](255) NULL,
    [PACity] [nvarchar](255) NULL,
    [PAZipCode] [nvarchar](255) NULL,
    [PACountry] [nvarchar](255) NULL,
    [PAState] [nvarchar](255) NULL,
    [IAStreet] [nvarchar](255) NULL,
    [IASuite] [nvarchar](255) NULL,
    [IACity] [nvarchar](255) NULL,
    [IAZipCode] [nvarchar](255) NULL,
    [IACountry] [nvarchar](255) NULL,
    [IAState] [nvarchar](255) NULL,
    [EAStreet] [nvarchar](255) NULL,
    [EASuite] [nvarchar](255) NULL,
    [EACity] [nvarchar](255) NULL,
    [EAZipCode] [nvarchar](255) NULL,
    [EACountry] [nvarchar](255) NULL,
    [EAState] [nvarchar](255) NULL,
    [PYAStreet] [nvarchar](255) NULL,
    [PYASuite] [nvarchar](255) NULL,
    [PYACity] [nvarchar](255) NULL,
    [PYAZipCode] [nvarchar](255) NULL,
    [PYACountry] [nvarchar](255) NULL,
    [PYAState] [nvarchar](255) NULL,
 CONSTRAINT [PK_AMA] PRIMARY KEY CLUSTERED 
(
    [AMAId] ASC
))

    Public Class AmaStatementEntity

        Public Sub New()
            _amaId = Guid.NewGuid
        End Sub
'Other properties not relavent to discussion removed

        <Nested(ColumnPrefix:="FL")))>
        Public Property FacilityLocation() As AddressEntity

        <Nested(ColumnPrefix:="BA"))>
        Public Property BillingAddress() As AddressEntity

        <Nested(ColumnPrefix:="PA"))>
        Public Property PatientAddress() As AddressEntity

        <Nested(ColumnPrefix:="IA"))>
        Public Property InsuredAddress() As AddressEntity

        <Nested(ColumnPrefix:="EA"))>
        Public Property EmployerAddress() As AddressEntity

        <Nested(ColumnPrefix:="PYA"))>
        Public Property PayorAddress() As AddressEntity

        Public Property AMAId() As Guid
    End Class

Public Class AddressEntity
    Public Property Street() As String
    Public Property Suite() As String
    Public Property City() As String
    Public Property State() As String
    Public Property ZipCode() As String
    Public Property Country() As String
    Public Property AddressId() As Guid
End Class
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1 回答 1

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一旦我找到了正确的术语，我就能找到答案：

实体框架 - 重用复杂类型

对于我的特定场景，这是你将如何做到的（示例显示了复杂类型引用之一的映射）：

        modelBuilder.ComplexType<AddressEntity>();
        modelBuilder.Entity<AmaStatementEntity>().ToTable("vAMA");
        modelBuilder.Entity<AmaStatementEntity>().HasKey(a => a.AMAId);
        modelBuilder.Entity<AmaStatementEntity>().Property(a => a.FacilityLocation.Street).HasColumnName("FLStreet");
        modelBuilder.Entity<AmaStatementEntity>().Property(a => a.FacilityLocation.Suite).HasColumnName("FLSuite");
        modelBuilder.Entity<AmaStatementEntity>().Property(a => a.FacilityLocation.City).HasColumnName("FLCity");
        modelBuilder.Entity<AmaStatementEntity>().Property(a => a.FacilityLocation.ZipCode).HasColumnName("FLZipCode");
        modelBuilder.Entity<AmaStatementEntity>().Property(a => a.FacilityLocation.Country).HasColumnName("FLCountry");
        modelBuilder.Entity<AmaStatementEntity>().Property(a => a.FacilityLocation.State).HasColumnName("FLState");
于 2013-05-16T13:15:11.513 回答