SQL查询问题。。。

Question

CREATE TABLE [dbo].[SupplementDemandClientConfirm](
[ID] [int] IDENTITY(1,1) NOT NULL,
[SupplementDemandID] [int] NULL,
[DemandID] [int] NULL,
[TransName] [nvarchar](10) COLLATE Chinese_PRC_CI_AS NULL,
[TransDate] [nvarchar](10) COLLATE Chinese_PRC_CI_AS NULL,
[CreateUser] [int] NULL,
[CreateTime] [datetime] NULL,
 CONSTRAINT [PK_SUPPLEMENTDEMANDCLIENTCONFI] PRIMARY KEY CLUSTERED 
(
[ID] ASC
)WITH (IGNORE_DUP_KEY = OFF) ON [PRIMARY]
) ON [PRIMARY]
以上是我的表结构，我想查询的是，与当前时间最接近的那条记录，请问各位这个要怎么查？

Answer 1

select *
from SupplementDemandClientConfirm t
where id = (select top 1 id from SupplementDemandClientConfirm where 分组列 = t.分组列 order by CreateTime desc)

Answer 2

select top 1 * from SupplementDemandClientConfirm 
order by CreateTime desc

Answer 3

select *
from SupplementDemandClientConfirm t
where CreateTime= (select max(CreateTime) from SupplementDemandClientConfirm where 分组列 = t.分组列)

Answer 4

select top 1 * from SupplementDemandClientConfirm order by abs(datediff(m,getdate(),CreateTime))

Answer 5

select top 1 * from SupplementDemandClientConfirm order by abs(datediff(ss,getdate(),CreateTime))
用秒来比.

Answer 6

select top(1) * from tb order by CreateTime desc

Answer 7

select top 1 * from SupplementDemandClientConfirm 
order by abs(datediff(mi,CreateTime,getdate()))