EF中的GUID COMB策略

Question

有没有办法使用CodeFirst设计为新的Entity Framework 4.1中的对象实现Guid COMB身份策略？我认为设置StoreGeneratedPattern会工作,但它仍然给我正常的GUID.

Answer 1

为什么要担心数据库中Guid列的默认值呢？为什么不像其他任何值一样在客户端上生成Guid.这要求您在客户端代码中有一个方法,它将生成类似COMB的guid:

public static Guid NewGuid()
{
    var guidBinary = new byte[16];
    Array.Copy( Guid.NewGuid().ToByteArray(), 0, guidBinary, 0, 8 );
    Array.Copy( BitConverter.GetBytes( DateTime.Now.Ticks ), 0, guidBinary, 8, 8 );
    return new Guid( guidBinary );
}

Guid的一个优点是,您可以在客户端生成它们而无需往返数据库.

Answer 2

我猜你使用SQL服务器作为你的数据库.这是不同MS工具之间不一致的一个很好的例子.SQL Server团队不建议将列newid()用作UNIQUEIDENTIFIER列的默认值,如果Guid在数据库中将property 指定为autogenerated,则ADO.NET团队会使用它.他们应该使用newsequentialid()!

如果你想要数据库生成的顺序Guids,你必须修改生成的表,它真的很复杂,因为你必须找到自动生成的默认约束,删除它并创建新的约束.这一切都可以在自定义数据库初始化程序中完成.这里有我的示例代码:

class Program
{

    static void Main(string[] args)
    {
        Database.SetInitializer(new CustomInitializer());
        using (var context = new Context())
        {
            context.TestEntities.Add(new TestEntity() { Name = "A" });
            context.TestEntities.Add(new TestEntity() { Name = "B" });
            context.SaveChanges();
        }
    }
}

public class CustomInitializer : DropCreateDatabaseAlways<Context>
{
    protected override void Seed(Context context)
    {
        base.Seed(context);

        context.Database.ExecuteSqlCommand(@"
            DECLARE @Name VARCHAR(100)

            SELECT @Name = O.Name FROM sys.objects AS O
            INNER JOIN sys.tables AS T ON O.parent_object_id = T.object_id
            WHERE O.type_desc LIKE 'DEFAULT_CONSTRAINT' 
              AND O.Name LIKE 'DF__TestEntities__Id__%'
              AND T.Name = 'TestEntities'

            DECLARE @Sql NVARCHAR(2000) = 'ALTER TABLE TestEntities DROP Constraint ' + @Name

            EXEC sp_executesql @Sql

            ALTER TABLE TestEntities
            ADD CONSTRAINT IdDef DEFAULT NEWSEQUENTIALID() FOR Id");
    }
}

public class TestEntity
{
    public Guid Id { get; set; }
    public string Name { get; set; }
}

public class Context : DbContext
{
    public DbSet<TestEntity> TestEntities { get; set; }

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

        modelBuilder.Entity<TestEntity>()
            .Property(e => e.Id)
            .HasDatabaseGeneratedOption(DatabaseGeneratedOption.Identity);
    }
}