日期:2014-05-18  浏览次数:20481 次

一个SQL语句,大家看看怎么写
如: 
  ID Name
0HS30014202011110801 300新金融(广发)_20111108
0HS30014202011111001 300新金融(广发)_20111110
0HS30016002011110102 300非金融_20111101  
0HS30016002011122101 300非金融_20111221
0HS30016002011122301 300非金融_20111223
0JUNBO17222011122001 大盘A_20111220
0JUNBO17222011122101 大盘A_20111221

ID的前10为是分类,同一组,后几位是区分,每天都会更新不同的组合,现在想取每个分类下,时间最新的一条数据,根据要求,取出的数据应该是
0HS30014202011111001 300新金融(广发)_20111110
0HS30016002011122301 300非金融_20111223
0JUNBO17222011122101 大盘A_20111221

(说明20111221日期,01代码今天第一次更新),这个SQL如何写

------解决方案--------------------
SQL code

create table tb
(
  ID varchar(50),
  NAME nvarchar(50)
)

insert into tb 
  
select '0HS30014202011110801', '300新金融(广发)_20111108' union all
select '0HS30014202011111001', '300新金融(广发)_20111110' union all
select '0HS30016002011110102', '300非金融_20111101'  union all
select '0HS30016002011122101', '300非金融_20111221' union all
select '0HS30016002011122301', '300非金融_20111223'  union all 
select '0JUNBO17222011122001', '大盘A_20111220' union all
select '0JUNBO17222011122101', '大盘A_20111221' 


;with cte as 
(
  select *,rownum=ROW_NUMBER ()over(partition by left(ID,10) order by SUBSTRING(id,10 + 1,LEN(id)-1) desc) from tb
)

select ID,NAME  from cte where rownum =1

ID                                                 NAME
-------------------------------------------------- --------------------------------------------------
0HS30014202011111001                               300新金融(广发)_20111110
0HS30016002011122301                               300非金融_20111223
0JUNBO17222011122101                               大盘A_20111221

------解决方案--------------------
SQL code

--处理表重复记录(查询和删除)
/******************************************************************************************************************************************************
1、Num、Name相同的重复值记录,没有大小关系只保留一条
2、Name相同,ID有大小关系时,保留大或小其中一个记录
整理人:中国风(Roy)

日期:2008.06.06
******************************************************************************************************************************************************/

--1、用于查询重复处理记录(如果列没有大小关系时2000用生成自增列和临时表处理,SQL2005用row_number函数处理)

--> --> (Roy)生成測試數據
 
if not object_id('Tempdb..#T') is null
    drop table #T
Go
Create table #T([ID] int,[Name] nvarchar(1),[Memo] nvarchar(2))
Insert #T
select 1,N'A',N'A1' union all
select 2,N'A',N'A2' union all
select 3,N'A',N'A3' union all
select 4,N'B',N'B1' union all
select 5,N'B',N'B2'
Go


--I、Name相同ID最小的记录(推荐用1,2,3),方法3在SQl05时,效率高于1、2
方法1:
Select * from #T a where not exists(select 1 from #T where Name=a.Name and ID<a.ID)

方法2:
select a.* from #T a join (select min(ID)ID,Name from #T group by Name) b on a.Name=b.Name and a.ID=b.ID

方法3:
select * from #T a where ID=(select min(ID) from #T where Name=a.Name)

方法4:
select a.* from #T a join #T b on a.Name=b.Name and a.ID>=b.ID group by a.ID,a.Name,a.Memo having count(1)=1 

方法5:
select * from #T a group by ID,Name,Memo having ID=(select min(ID)from #T where Name=a.Name)

方法6:
select * from #T a where (select count(1) from #T where Name=a.Name and ID<a.ID)=0

方法7:
select * from #T a where ID=(select top 1 ID from #T where Name=a.name order by ID)

方法8:
select * from #T a where ID!>all(select ID from #T where Name=a.Name)

方法9(注:ID为唯一时可用):
select * from #T a where ID in(select min(ID) from #T group by Name)

--SQL2005:

方法10:
select ID,Name,Memo from (select *,min(ID)over(partition by Name) as MinID from #T a)T where ID=MinID

方法11:

select ID,Name,Memo from (select *,row_number()over(partition by Name order by ID) as MinID from #T a)T where MinID=1

生成结果:
/*
ID          Name Memo
----------- ---- ----
1           A    A1
4           B    B1

(2 行受影响)
*/


--II、Name相同ID最大的记录,与min相反:
方法1:
Select * from #T a where not exists(select 1 from #T where