请问SQL中如何GROUP BY 与TOP

一个表中的数据

ID nickname addtime
1 a 2011-12-19 11:11:12
2 b 2011-12-19 11:11:12
1 c 2011-12-19 11:11:10
2 c 2011-12-19 11:11:10
1 a 2011-12-19 11:11:11
2 b 2011-12-19 11:11:11

需求是根据时间倒序，取出每个ID，nickname 的前两条数据。

SQL code

;
WITH    tmp
          AS ( SELECT   * ,
                        rn = ROW_NUMBER() OVER ( PARTITION BY id ORDER BY nickname )
               FROM     tb
             )
    SELECT  *
    FROM    tmp
    WHERE   rn <= 2

SQL code

select * from 
(select no=row_number() over(partition by id order by addtime desc),* from tb)t
 where no<3

SQL code

select 
  distinct b.*
from
  tb a
cross apply
  (select top 2 * from tb where id=a.id order by addtime desc)b

SQL code

select distinct b.* from tb a
cross apply(select top 2 * from tb where id=a.id order by addtime desc) b

SQL code

if object_id('tb') is not null
   drop table tb
go
create table tb
(
 id int,
 nickname varchar(10),
 addtime datetime
)
go
insert into tb
select 1,'a','2011-12-19 11:11:12' union all
select 2,'b','2011-12-19 11:11:12' union all
select 1,'c','2011-12-19 11:11:10' union all
select 2,'c','2011-12-19 11:11:10' union all
select 1,'a','2011-12-19 11:11:11' union all
select 2,'b','2011-12-19 11:11:11'
go
select id,nickname,addtime from 
(
 select *,row=row_number() over(partition by id,nickname order by getdate()) from tb
) t where row between 1 and 2
go
/*
id          nickname   addtime
----------- ---------- -----------------------
1           a          2011-12-19 11:11:12.000
1           a          2011-12-19 11:11:11.000
1           c          2011-12-19 11:11:10.000
2           b          2011-12-19 11:11:12.000
2           b          2011-12-19 11:11:11.000
2           c          2011-12-19 11:11:10.000

(6 行受影响)
*/

SQL code

---------------------------------
--  Author: liangCK 小梁
--  Title : 查每个分组前N条记录
--  Date  : 2008-11-13 17:19:23
---------------------------------

--> 生成测试数据: #T
IF OBJECT_ID('tempdb.dbo.#T') IS NOT NULL DROP TABLE #T
CREATE TABLE #T (ID VARCHAR(3),GID INT,Author VARCHAR(29),Title VARCHAR(39),Date DATETIME)
INSERT INTO #T
SELECT '001',1,'邹建','深入浅出SQLServer2005开发管理与应用实例','2008-05-10' UNION ALL
SELECT '002',1,'胡百敬','SQLServer2005性能调校','2008-03-22' UNION ALL
SELECT '003',1,'格罗夫Groff.J.R.','SQL完全手册','2009-07-01' UNION ALL
SELECT '004',1,'KalenDelaney','SQLServer2005技术内幕存储引擎','2008-08-01' UNION ALL
SELECT '005',2,'Alex.Kriegel.Boris.M.Trukhnov','SQL宝典','2007-10-05' UNION ALL
SELECT '006',2,'飞思科技产品研发中心','SQLServer2000高级管理与开发','2007-09-10' UNION ALL
SELECT '007',2,'胡百敬','SQLServer2005数据库开发详解','2008-06-15' UNION ALL
SELECT '008',3,'陈浩奎','SQLServer2000存储过程与XML编程','2005-09-01' UNION ALL
SELECT '009',3,'赵松涛','SQLServer2005系统管理实录','2008-10-01' UNION ALL
SELECT '010',3,'黄占涛','SQL技术手册','2006-01-01'

--SQL查询如下:

--按GID分组,查每个分组中Date最新的前2条记录


--1.字段ID唯一时:
SELECT * FROM #T AS T WHERE ID IN(SELECT TOP 2 ID FROM #T WHERE GID=T.GID ORDER BY Date DESC)

--2.如果ID不唯一时:
SELECT * FROM #T AS T WHERE 2>(SELECT COUNT(*) FROM #T WHERE GID=T.GID AND Date>T.Date)

--SQL Server 2005 使用新方法

--3.使用ROW_NUMBER()进行排位分组
SELECT ID,GID,Author,Title,Date
FROM
(
   SELECT rid=ROW_NUMBER() OVER(PARTITION BY GID ORDER BY Date DESC),*
   FROM #T
) AS T
WHERE rid<=2

--4.使用APPLY
SELECT DISTINCT b.*
FROM #T AS a
CROSS APPLY
(
    SELECT TOP(2) * FROM #T WHERE a.GID=GID ORDER BY Date DESC
) AS b


--结果
/*

ID   GID         Author                        Title                                   Date
---- ----------- ----------------------------- --------------------------------------- -----------------------
003  1           格罗夫Groff.J.R.                 SQL完全手册                                 2009-07-01 00:00:00.000
004  1           KalenDelaney                  SQLServer2005技术内幕存储引擎                   2008-08-01 00:00:00.000
005  2           Alex.Kriegel.Boris.M.Trukhnov SQL宝典                                   2007-10-05 00:00:00.000
007  2           胡百敬                           SQLServer2005数据库开发详解                    2008-06-15 00:00:00.000
009  3           赵松涛                           SQLServer2005系统管理实录                     2008-10-01 00:00:00.000
010  3           黄占涛                           SQL技术手册                                 2006-01-01 00:00:00.000

(6 行受影响)
*/

select 
  distinct b.*
from
  tb a
cross apply
  (select top 2 * from tb where id=a.id order by addtime desc)b

F姐姐依然那么犀利

SQL code

--应该按倒序排
if object_id('tb') is not null
   drop table tb
go
create table tb
(
 id int,
 nickname varchar(10),
 addtime datetime
)
go
insert into tb
select 1,'a','2011-12-19 11:11:12' union all
select 2,'b','2011-12-19 11:11:12' union all
select 1,'c','2011-12-19 11:11:10' union all
select 2,'c','2011-12-19 11:11:10' union all
select 1,'a','2011-12-19 11:11:11' union all
select 2,'b','2011-12-19 11:11:11'
go
select id,nickname,addtime from 
(
 select *,row=row_number() over(partition by id,nickname order by addtime desc) from tb
) t where row between 1 and 2
go
/*
id          nickname   addtime
----------- ---------- -----------------------
1           a          2011-12-19 11:11:12.000
1           a          2011-12-19 11:11:11.000
1           c          2011-12-19 11:11:10.000
2           b          2011-12-19 11:11:12.000
2           b          2011-12-19 11:11:11.000
2           c          2011-12-19 11:11:10.000

(6 行受影响)
*/