编纂sql语句，查询除了星期六星期日最长连续的数据

select datepart(W,GETDATE())/*  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'  ;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 name,start_day,end_day,logindays-(select COUNT(1) from [tbl] c where c.name=a.name and c.logindate between a.start_day and a.end_day and DATEPART(W,c.logindate)=1 or  DATEPART(W,c.logindate)=7)+1 as logindaysfrom m a   where logindays in(select MAX(logindays) from m b   where a.name=b.name)  /*  name    start_day    end_day    logindaysa1    2011-01-02    2011-01-04    2a2    2011-01-10    2011-01-11    2a2    2011-01-07    2011-01-08    2*/  顺便改了一下