为什么EF 4不会为具有唯一索引的列生成FK关联关联？

Question

我遇到过一个场景,其中Entity Framework 4.0没有生成与具有唯一索引的表支持的实体的关联,我想知道为什么.

基本设置是这样的:假设我在SQL Server 2008 R2中有两个表和一个外键关系:

CREATE TABLE [dbo].[User](
    [Id] [int] IDENTITY(1,1) NOT NULL,
    [GroupId] [int] NULL,
 CONSTRAINT [PK_User] PRIMARY KEY CLUSTERED 
(
    [Id] ASC
)WITH (PAD_INDEX  = OFF, STATISTICS_NORECOMPUTE  = OFF, IGNORE_DUP_KEY = OFF,
ALLOW_ROW_LOCKS  = ON, ALLOW_PAGE_LOCKS  = ON) ON [PRIMARY]
) ON [PRIMARY]

CREATE TABLE [dbo].[Group](
[Id] [int] IDENTITY(1,1) NOT NULL,
 CONSTRAINT [PK_Group] PRIMARY KEY CLUSTERED 
(
    [Id] ASC
)WITH (PAD_INDEX  = OFF, STATISTICS_NORECOMPUTE  = OFF, IGNORE_DUP_KEY = OFF,
    ALLOW_ROW_LOCKS  = ON, ALLOW_PAGE_LOCKS  = ON) ON [PRIMARY]
) ON [PRIMARY]

ALTER TABLE [dbo].[User]  WITH CHECK ADD  CONSTRAINT [FK_User_Group] 
    FOREIGN KEY([GroupId])
REFERENCES [dbo].[Group] ([Id])

此外,假设存在以下索引:

CREATE NONCLUSTERED INDEX [IX_Group] ON [dbo].[Group] 
(
[Id] ASC
)WITH (PAD_INDEX  = OFF, STATISTICS_NORECOMPUTE  = OFF, SORT_IN_TEMPDB = OFF, IGNORE_DUP_KEY = OFF, DROP_EXISTING = OFF, ONLINE = OFF, ALLOW_ROW_LOCKS  = ON, ALLOW_PAGE_LOCKS  = ON) ON [PRIMARY]

如果我告诉在Visual Studio 2010设计者生成ADO.NET实体数据模型我得到两个类的模型,User并且Group,User有称为导航属性Group.这一切都很好,很好.

现在,让我们说索引看起来像这样:

CREATE UNIQUE NONCLUSTERED INDEX [IX_Group] ON [dbo].[Group] 
(
    [Id] ASC
)WITH (PAD_INDEX  = OFF, STATISTICS_NORECOMPUTE  = OFF, SORT_IN_TEMPDB = OFF, IGNORE_DUP_KEY = OFF, DROP_EXISTING = OFF, ONLINE = OFF, ALLOW_ROW_LOCKS  = ON, ALLOW_PAGE_LOCKS  = ON) ON [PRIMARY]

That is, the only thing I have done is make the index a unique index. Having done this, when I tell Visual Studio's designer to generate an Entity Model, the association between users and groups doesn't show up and the User has no navigation properties. Inspecting the generated EDMX file reveals that the storage model has no AssociationSet at all.

Can anyone explain why this is? Why does the unique index prevent EF from modeling the relationship?

Thank you.

Answer 1

唯一索引允许1 NULL值,主键不允许NULLS.当没有任何东西等于NULL而不是另一个NULL时,你将如何匹配NULL