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如何连续查询天数

更新时间:2012-4-7:  来源:51代码网

create table #loginlog(logintime datetime,u_id int)

insert into #loginlog select '2011-12-16',907
insert into #loginlog select '2011-12-17',907
insert into #loginlog select '2011-12-18',907
insert into #loginlog select '2011-12-14',1100
insert into #loginlog select '2011-12-15',1100
insert into #loginlog select '2011-12-16',1100
insert into #loginlog select '2011-12-13',1200
insert into #loginlog select '2011-12-15',1200
insert into #loginlog select '2011-12-16',1200
insert into #loginlog select '2011-12-17',1200
insert into #loginlog select '2011-12-18',1200
insert into #loginlog select '2011-12-19',1200
insert into #loginlog select '2011-12-20',1200
insert into #loginlog select '2011-12-21',1200
insert into #loginlog select '2011-12-21',1200
insert into #loginlog select '2011-12-22',1200
insert into #loginlog select '2011-12-23',1200
insert into #loginlog select '2011-12-24',1200
insert into #loginlog select '2011-12-25',1200
insert into #loginlog select '2011-12-26',1200
insert into #loginlog select '2011-12-27',1200
insert into #loginlog select '2011-12-28',1200
go
--同一个帐号,连续登陆的最大天数
;WITH cte AS
(
  SELECT b.u_id,CONVERT(CHAR(8),DATEADD(day,-sn,logintime),112)AS dn,COUNT(1) AS counts
  FROM 
  (
  SELECT u_id,ROW_NUMBER()OVER(PARTITION BY u_id ORDER BY logintime) AS sn,logintime
  FROM #loginlog
  )AS b
  GROUP BY b.u_id,CONVERT(CHAR(8),DATEADD(day,-sn,b.logintime),112)
)
SELECT cte.u_id,MAX(counts) as counts FROM cte GROUP BY cte.u_id

这是查询账号连续几天登陆次数的
其中结果账号1200:显示9
事实上账号1200:应该显示是14
其中
insert into #loginlog select '2011-12-21',1200
insert into #loginlog select '2011-12-21',1200
有2条,是该用用户这天登陆了2次,如果'2011-12-21',1200数据只有一条的话,那结果就是正确的显示14
请问该怎么改才能支持有每天不断重复登陆的,麻烦大家了!!
将ROW_NUMBER改为 dense_rank

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