如何求取一个月中连续登陆天数最大值问题，不使用游标怎么解决。有一个思路是：某天登陆记为1，没有登陆记为0.这样一个月中就会出现111010。。。。。这样的形式，不过后面我就不知道怎么写SQL了。
测试数据如下：
create table LoginTable (name varchar(10),logindate datetime)
insert into LoginTable values('A','2011-03-07'),
('A','2011-03-1'),
('A','2011-03-2'),
('A','2011-03-3'),
('A','2011-03-6'),
('A','2011-03-8'),
('A','2011-03-10'),
('A','2011-03-11'),
('A','2011-03-12'),
('A','2011-03-15'),
('A','2011-03-16'),
('A','2011-03-20'),
('A','2011-03-25'),
('A','2011-03-29');
------解决方案--------------------
- SQL code
create table LoginTable (name varchar(10),logindate date)insert into LoginTable values('A','2011-03-07'),('A','2011-03-1'),('A','2011-03-2'),('A','2011-03-3'),('A','2011-03-6'),('A','2011-03-8'),('A','2011-03-10'),('A','2011-03-11'),('A','2011-03-12'),('A','2011-03-15'),('A','2011-03-16'),('A','2011-03-20'),('A','2011-03-25'),('A','2011-03-29');with d as(select logindate,(select min(b.logindate) from LoginTable b where b.logindate>=a.logindateand not exists (select * from LoginTable c where c.logindate=dateadd(dd,1,b.logindate))) as grpfrom LoginTable a),m as(select min(logindate) as start_range,max(logindate) as end_rangefrom d group by grp)select max(DATEDIFF(DD,start_range,end_range)+1) as maxday from mwhere DATEDIFF(DD,start_range,end_range)<>0/*maxday3*/
------解决方案--------------------