如何查询出连续登陆的最长天数

最近遇到不少关于解决连续登陆天数的帖子。这类问题实际也就是我们经常遇到的孤岛问题的一个变种

解决这类问题，我们有一个最基本的思路：

step 1:找出间断之后的点，为他们分配行号（这是孤岛的起点）

step 2:找出间断之前的点，为他们分配行号（这是孤岛的终点）

step 3:以行号相等作为条件，匹配孤岛的起点和终点

在实现以上三步后，我们基本上就能解决这类问题了，一下我以三种方法演示：

/*

name logindate

a1 2011-1-2

a1 2011-1-3

a1 2011-1-4

a1 2011-1-7

a1 2011-1-12

a1 2011-1-13

a1 2011-1-16

a2 2011-1-7

a2 2011-1-8

a2 2011-1-10

a2 2011-1-11

a2 2011-1-13

a2 2011-1-24

---------------------------------------------

我需要的结果是：

name start_day end_day logindays

a1 2011-1-2 2011-1-4 3

a2 2011-1-7 2011-1-8 2

a2 2011-1-10 2011-1-11 2

*/

--> 测试数据：[tbl]

if object_id('[tbl]') is not null drop table [tbl]

create table [tbl]([name] varchar(2),[logindate] date)

insert [tbl]

select 'a1','2011-1-2' union all

select 'a1','2011-1-3' union all

select 'a1','2011-1-4' union all

select 'a1','2011-1-7' union all

select 'a1','2011-1-12' union all

select 'a1','2011-1-13' union all

select 'a1','2011-1-16' union all

select 'a2','2011-1-7' union all

select 'a2','2011-1-8' union all

select 'a2','2011-1-10' union all

select 'a2','2011-1-11' union all

select 'a2','2011-1-13' union all

select 'a2','2011-1-24'

--方法1

;with t as(

select [name],[logindate],

(select min(b.[logindate]) from tbl b

where b.[logindate]>=a.[logindate] and b.name=a.name

and not exists (select * from tbl c

where c.[logindate]=dateadd(dd,1,b.[logindate]) and c.name=b.name)) as grp

from tbl a

),m

as(

select [name],min([logindate]) as start_day,max(grp) as end_day

from t group by grp,name

)

select *,(datediff(dd,start_day,end_day)+1) as logindays from m a

where (datediff(dd,start_day,end_day)+1) in(

select max(datediff(dd,start_day,end_day)+1) from m b

where a.name=b.name)

---------------------------------------------------------------------------

---------------------------------------------------------------------------

--方法2

declare @date datetime

select @date = min(logindate) from tbl

;with ach as

(

select [name],logindate,

id=row_number() over (partition by [name] order by logindate)

from tbl

),

t as(

select [name],min(logindate) mindate,max(logindate) maxdate,

(datediff(dd,min(logindate),max(logindate))+1) dddate

from ach

group by [name],datediff(dd,@date,logindate)-id

--order by [name],mindate

)

select * from t a where dddate in(select max(dddate) from t b where

a.name=b.name)

--------------------------------------------------------------------------

--------------------------------------------------------------------------

--方法3

;with t as

(

select name,[logindate],dateadd(dd,

-row_number()over(partition by name

order by [logindate]),[logindate]) as diff from tbl

),

m as(

select name,min([logindate]) as start_day,max([logindate]) as end_day,

(datediff(dd,min([logindate]),max([logindate]))+1) as logindays

from t

group by name,diff

)

select * from m a

where logindays in(select MAX(logindays) from m b

where a.name=b.name)

/*

name start_day end_day logindays

a1 2011-01-02 2011-01-04 3

a2 2011-01-10 2011-01-11 2

a2 2011-01-07 2011-01-08 2

*/

关于连续登陆问题也就是时间孤岛问题，所以我们的解决思路就等同于求孤岛。

在学习的过程中我建议大家能抓住最基本的方法，不要一味的追求最简单的方法。

任何一个简单的方法都是建立在一定得基础知识和熟练程度上的，只有当你熟练

掌握了最基本的东西才可能去发现更简单的办法。

来自：http://blog.csdn.net/travylee/article/details/7460030