现在数据库里边有时间time、位置local和计数器字段counter现在要根据时间进行汇总,汇总后counter的值是求最新的一条
请问怎么写这样的sql
例如数据库里边数据为
time local counter
2009-7-25 1 100
2009-7-25 2 200
2009-7-25 3 300
2009-7-25 4 400
2009-7-26 1 100
2009-7-26 2 200
2009-7-26 3 300
2009-7-26 4 500根据sql求出的结果为
2009-7-25 4 400
2009-7-26 4 500
请问大家如何编写?
请问怎么写这样的sql
例如数据库里边数据为
time local counter
2009-7-25 1 100
2009-7-25 2 200
2009-7-25 3 300
2009-7-25 4 400
2009-7-26 1 100
2009-7-26 2 200
2009-7-26 3 300
2009-7-26 4 500根据sql求出的结果为
2009-7-25 4 400
2009-7-26 4 500
请问大家如何编写?
where counter=(select max(counter) from tb
where a.time=time)
not exists
(select 1 from tb where [time]=t.[time] and counter>t.counter)
/******************************************************************************************************************************************************
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
--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 Name=a.Name and ID>a.ID)方法2:
select a.* from #T a join (select max(ID)ID,Name from #T group by Name) b on a.Name=b.Name and a.ID=b.ID order by ID方法3:
select * from #T a where ID=(select max(ID) from #T where Name=a.Name) order by ID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 max(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 desc)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 max(ID) from #T group by Name)--SQL2005:方法10:
select ID,Name,Memo from (select *,max(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 desc) as MinID from #T a)T where MinID=1生成结果2:
/*
ID Name Memo
----------- ---- ----
3 A A3
5 B B2(2 行受影响)
*/
not exists(select 1 from tb where [time]=t.[time] and counter>t.counter)or
select * from tb t where
not exists(select 1 from tb where [time]=t.[time] and local>t.local)
where counter=(select max(counter) from tb where [time]=a.[time])
counter=(select top 1 counter from tb where [time]=t.[time] order by counter desc)
--or
select * from tb t where
local=(select top 1 local from tb where [time]=t.[time] order by local desc)
--同一个时间,local最大,入库最晚吧?
select * from tb a
where not exists(select 1 from tb where [time]=a.[time] and [local]>a.[local])
select max(counter) keep(DENSE_RANK last ORDER BY COLLECTTIME asc) from table group by function(collectime)这样汇总的结果是把collectime的数据汇总到一小时,而求出的counter是这个小时的最后一条入库数据。oracle有这样的函数,而sqlserve还没有这样的方法,不知道如何处理,请大家帮忙想想
select * from tb as a
where not exists(select 1 from tb where [time]=a.[time] and [local]>a.[local])