SQL分组获取第一条记录,谁能讲解下?

[火星大能猫]

表数据如图

eqpid为设备编号,boottime为开机时间,id为自增长,shuttime字段无用.
一个设备会有多次开机记录,需要获取每个设备的第一次开机记录

SELECT eqpId,bsu1.bootTime FROM BootStartUp bsu1
WHERE NOT EXISTS
(
	SELECT * FROM 
BootStartUp bsu2 WHERE bsu1.eqpId=bsu2.eqpId 
AND bsu1.bootTime> bsu2.bootTime
	)ORDER BY bsu1.bootTime

根据网上比较普遍的做法是应该这样.
但对于这样的表连接查询不是很明白原理,谁能仔细解释下,或者有没有其它更容易理解的做法?

[叶子]

--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 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)
*/

[x-code-man]

上面的sql写得很棒啊

[x-code-man]

我说的是楼主的

[zou_yu_lin]

select eqpid,min(boottime) from BootStartUp group by eqpid

[火星大能猫]

[Quote=引用 4 楼 的回复:]

select eqpid,min(boottime) from BootStartUp group by eqpid
[/Quote]
居然忘记min了!!!!!

[火星大能猫]

谁能讲解下SELECT * FROM
BootStartUp bsu2 WHERE bsu1.eqpId=bsu2.eqpId
AND bsu1.bootTime> bsu2.bootTime的主要意义?

[liangjingji]

[Quote=引用 6 楼 的回复:]

谁能讲解下SELECT * FROM
BootStartUp bsu2 WHERE bsu1.eqpId=bsu2.eqpId
AND bsu1.bootTime> bsu2.bootTime的主要意义?
[/Quote]就把表BootStartUp的列bootTime在eqpId 相等的时候选出那个相对较大的bootTime的值.

[liangjingji]

他的意义就是一个筛选的作用,把比较的晚的时间个选出来