如何插入使用顺序 GUID 作为主键的 table？

Question

这是我正在查看的 table 的简化版本：

CREATE TABLE [dbo].[FrustratingTable]
(
    [Id] Uniqueidentifier NOT NULL
    , [SecondField] [datetime]
    , [ThirdField] varchar(128)
)

我想在此 table 中插入新记录。我尝试了 3 种方法：

INSERT INTO [dbo].[FrustratingTable] (Id, SecondField, ThirdField)
    SELECT newid() as Id, 
           '6/25/2015' as SecondField, 'Example' as ThirdField

这种方法会插入，但生成的键不像 table

中的其他键那样是一个很好的顺序 GUID

INSERT INTO [dbo].[FrustratingTable] (Id, SecondField, ThirdField)
SELECT NEWSEQUENTIALID() as Id, '6/25/2015' as SecondField, 'Example' as ThirdField

失败并出现错误

The newsequentialid() built-in function can only be used in a DEFAULT expression for a column of type 'uniqueidentifier' in a CREATE TABLE or ALTER TABLE statement. It cannot be combined with other operators to form a complex scalar expression.

INSERT INTO [dbo].[FrustratingTable] (SecondField,ThirdField)
SELECT '6/25/2015' as SecondField, 'Example' as ThirdField

失败并出现错误

Cannot insert the value NULL into column 'id', table 'mydatabase.dbo.frustratingtable'; column does not allow nulls. INSERT fails.

是否可以在不改变 table 定义的情况下解决这个问题？

Answer 1

ALTER TABLE FrustratingTable
ALTER COLUMN id uniqueidentifier not null default newsequentialid()

Answer 2

您可以通过使用 table 变量来做到这一点：

declare @t table (
    ID uniqueidentifier not null default newsequentialid(),
    SecondField datetime,
    ThirdField varchar(128)
)
insert into @t (SecondField,ThirdField)
    output inserted.ID,inserted.SecondField,inserted.ThirdField
    into FrustratingTable
values
('20150101','abc'),
('20150201','def'),
('20150301','ghi')

select * from FrustratingTable

结果：

Id                                   SecondField             ThirdField
------------------------------------ ----------------------- ------------
1FEBA239-091C-E511-9B2F-78ACC0C2596E 2015-01-01 00:00:00.000 abc
20EBA239-091C-E511-9B2F-78ACC0C2596E 2015-02-01 00:00:00.000 def
21EBA239-091C-E511-9B2F-78ACC0C2596E 2015-03-01 00:00:00.000 ghi

由于 table 变量通过 default 设置值，我们可以使用 NEWSEQUENTIALID().

当然，对于非常大的数据集，暂时让两个数据副本潜伏在周围是有代价的。

另一种方法是使用称为 COMB 的旧解决方案，它在引入 NEWSEQUENTIALID() 之前使用过：

SELECT CAST(CAST(NEWID() AS BINARY(10)) + CAST(GETDATE() AS BINARY(6)) AS UNIQUEIDENTIFIER)

生成比 NEWID() 本身具有更好局部性的 uniqueidentifiers。

Answer 3

好的，如果你首先从 this answer 中获取 [IncrementGuid] 函数，那么你可以这样做，

Fiddle Here

INSERT [dbo].[FrustratingTable]
SELECT
      [dbo].[IncrementGuid](MAX([Id])),
      '01/01/01',
      '3'
  FROM
      [dbo].[FrustratingTable];

警告：

一旦您查看了其他答案中的功能，您就会同意，一定有更好的方法。

更改需要 GUID 顺序的代码。

如何插入使用顺序 GUID 作为主键的 table？

How to INSERT into a table that uses sequential GUIDs as a primary key?

sql

guid

sql-server-2012

警告：