查询每个人最晚的纪要！求搭救

2013-09-09

查询每个人最晚的纪录！！求搭救表index userid workidtime1112013-07-23 12:38:10 2222013-07-23 12:38:11

查询每个人最晚的纪录！！求搭救
表
index userid workid       time
1      1      1     2013-07-23 12:38:10
2      2      2     2013-07-23 12:38:11
3      3      3     2013-07-23 12:38:12
4      1      4     2013-07-23 12:38:13
5      2      5     2013-07-23 12:38:14
6      3      6     2013-07-23 12:38:15

想要查出每个人最晚那条记录如
4      1      4     2013-07-23 12:38:13
5      2      5     2013-07-23 12:38:14
6      3      6     2013-07-23 12:38:15

求搭救 sql?查询?
[解决办法]
select * from 表 a
where time=(select max(time) from 表 where a.userid=userid)
[解决办法]
select a.*
from tb a
inner join (select userid,max(time) as time from tb group by userid)b
on a.userid=b.userid and a.time=b.time
[解决办法]

create table tb([index] int, userid int,workid int,[time] datetime)
insert into tb
select 1,1,1,'2013-07-23 12:38:10'
union all select 2,2,2,'2013-07-23 12:38:11'
union all select 3,3,3,'2013-07-23 12:38:12'
union all select 4,1,4,'2013-07-23 12:38:13'
union all select 5,2,5,'2013-07-23 12:38:14'

union all select 6,3,6,'2013-07-23 12:38:15'

select * from tb

select a.*
from tb a
inner join (select userid,max(time) as time from tb group by userid)b
on a.userid=b.userid and a.time=b.time
order by userid

drop table tb

/*
4142013-07-23 12:38:13.000
5252013-07-23 12:38:14.000
6362013-07-23 12:38:15.000
*/

[解决办法]
--> 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)
*/

[解决办法]
create table tb([index] int, userid int,workid int,[time] datetime)
insert into tb
select 1,1,1,'2013-07-23 12:38:10'
union all select 2,2,2,'2013-07-23 12:38:11'
union all select 3,3,3,'2013-07-23 12:38:12'

union all select 4,1,4,'2013-07-23 12:38:13'
union all select 5,2,5,'2013-07-23 12:38:14'
union all select 6,3,6,'2013-07-23 12:38:15'

WITH CTE AS
(SELECT ROW_NUMBER() OVER(PARTITION BY userid ORDER BY [time] DESC) AS [rank],* FROM tb
)
SELECT * FROM CTE WHERE [rank] =1

drop table tb

/*
rankindexuseridworkidtime
14142013-07-23 12:38:13.000
15252013-07-23 12:38:14.000
16362013-07-23 12:38:15.000
*/
[解决办法]
二次循环

第一循环：人员
——可以用user表
——也可以group by一下

第二循环：top 1 where 姓名=第一循环的姓名 order by time desc
[解决办法]

引用:

是50万条数据

if object_id('[TB]') is not null drop table [TB]
go
create table [TB] ([index] int,userid int,workid int,time datetime)
insert into [TB]
select 1,1,1,'2013-07-23 12:38:10' union all
select 2,2,2,'2013-07-23 12:38:11' union all
select 3,3,3,'2013-07-23 12:38:12' union all
select 4,1,4,'2013-07-23 12:38:13' union all
select 5,2,5,'2013-07-23 12:38:14' union all
select 6,3,6,'2013-07-23 12:38:15'

select * from [TB]


SELECT  *
FROM    dbo.TB
WHERE   NOT EXISTS ( SELECT 1      --注意使用not exists，最好匹配对应的索引，参考：http://blog.csdn.net/orchidcat/article/details/6267552
                     FROM   TB B
                     WHERE  TB.userid = B.userid
                            AND B.TIME > TB.TIME )

/*
indexuseridworkidtime
4142013-07-23 12:38:13.000
5252013-07-23 12:38:14.000
6362013-07-23 12:38:15.000*/

[解决办法]
2005 Row_Number + Over子句

[解决办法]
try this,


-- 建索引
create index idx_[表名]_userid_time on [表名](userid,[time])

-- 查询
select a.* 
 from [表名] a
 inner join 
 (select userid,
         max([time]) 'maxtime'
   from [表名]
   group by userid) b on a.userid=b.userid and a.[time]=b.maxtime

[解决办法]

引用:

表
index userid workid       time
1      1      1     2013-07-23 12:38:10
2      2      2     2013-07-23 12:38:11
3      3      3     2013-07-23 12:38:12
4      1      4     2013-07-23 12:38:13
5      2      5     2013-07-23 12:38:14
6      3      6     2013-07-23 12:38:15

想要查出每个人最晚那条记录如
4      1      4     2013-07-23 12:38:13
5      2      5     2013-07-23 12:38:14
6      3      6     2013-07-23 12:38:15

求搭救

IF OBJECT_ID('Ta') IS NOT NULL
DROP TABLE Ta
GO
CREATE TABLE Ta
( [index] INT,
userid INT,
workid INT,
[time] DATETIME
)
GO
INSERT Ta
SELECT 1,1,1,'2013-07-23 12:38:10' UNION
SELECT 2,2,2,'2013-07-23 12:38:11' UNION
SELECT 3,3,3,'2013-07-23 12:38:12' UNION
SELECT 4,1,4,'2013-07-23 12:38:13' UNION
SELECT 5,2,5,'2013-07-23 12:38:14' UNION
SELECT 6,3,6,'2013-07-23 12:38:15'

SELECT *
FROM TA a
WHERE EXISTS (

              SELECT 1
              FROM TA b
              WHERE A.userid=b.userid and a.[time]>b.[time]
              )

[解决办法]
12不正确!
[解决办法]
select * from
(select *,row_number() over(partition by userid order by time desc) as RowNO from table1) as a
where a.RowNO=1

[解决办法]

if object_id('[TB]') is not null drop table [TB]
go
create table [TB] ([index] int,userid int,workid int,time datetime)
insert into [TB]
select 1,1,1,'2013-07-23 12:38:10' union all
select 2,2,2,'2013-07-23 12:38:11' union all
select 3,3,3,'2013-07-23 12:38:12' union all
select 4,1,4,'2013-07-23 12:38:13' union all
select 5,2,5,'2013-07-23 12:38:14' union all
select 6,3,6,'2013-07-23 12:38:15'
 
--应该创建userid asc,time desc 复合索引
CREATE INDEX IX_TB_fieldlist ON dbo.[TB]
(
userid,
[time] DESC
) 

--50W数据，一般的服务器级别的硬件问题不大
SELECT b.*
FROM(SELECT DISTINCT userid FROM TB) A--不过你应该有现在的user表，用user表代替这行代码，效率会更好
CROSS APPLY
(SELECT TOP(1) * FROM TB M WHERE M.userid = A.userid ORDER BY [time] DESC) B
/*
indexuseridworkidtime
4142013-07-23 12:38:13.000
5252013-07-23 12:38:14.000
6362013-07-23 12:38:15.000
*/

[解决办法]

引用:

select * from tb

select a.*
from tb a
inner join (select userid,max(time) as time from tb group by userid)b
on a.userid=b.userid and a.time=b.time
order by userid

drop table tb

/*
4142013-07-23 12:38:13.000
5252013-07-23 12:38:14.000
6362013-07-23 12:38:15.000
*/

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