create table if not exists orde(id varchar(10),date datetime,orders varchar(10));
insert into orde values('1' , '2019/1/1',10 );
insert into orde values('1' , '2019/1/2',109 );
insert into orde values('1' , '2019/1/3',150 );
insert into orde values('1' , '2019/1/4',99);
insert into orde values('1' , '2019/1/5',145);
insert into orde values('1' , '2019/1/6',1455);
insert into orde values('1' , '2019/1/7',199);
insert into orde values('1' , '2019/1/8',188 );
insert into orde values('4' , '2019/1/1',10 );
insert into orde values('2' , '2019/1/2',109 );
insert into orde values('3' , '2019/1/3',150 );
insert into orde values('4' , '2019/1/4',99);
insert into orde values('5' , '2019/1/5',145);
insert into orde values('6' , '2019/1/6',1455);
insert into orde values('7' , '2019/1/7',199);
insert into orde values('8' , '2019/1/8',188 );
insert into orde values('9' , '2019/1/1',10 );
insert into orde values('9' , '2019/1/2',109 );
insert into orde values('9' , '2019/1/3',150 );
insert into orde values('9' , '2019/1/4',99);
insert into orde values('9' , '2019/1/6',145);
insert into orde values('9' , '2019/1/9',1455);
insert into orde values('9' , '2019/1/10',199);
insert into orde values('9' , '2019/1/13',188 );

查看數(shù)據(jù)表：

2.使用row_number() over() 排序函數(shù)計算每個id的排名，SQL如下：

select *,row_number() over(partition by id order by date ) 'rank'
from orde
where orders is not NULL;

查看數(shù)據(jù)表:

3.將date日期字段減去rank排名字段，SQL如下：

select *,DATE_SUB(a.date,interval a.rank day) 'date_sub'
from(
select *,row_number() over(partition by id order by date ) 'rank'
from orde
where orders is not NULL
) a;

查看數(shù)據(jù)：

4.根據(jù)id和date分組并計算分組后的數(shù)量(count)、計算最早登錄和最晚登錄的時間，SQL如下：

select b.id,min(date) 'start_time',max(date) 'end_time',count(*) 'date_count'
from(
select *,DATE_SUB(a.date,interval a.rank day) 'date_sub'
from(
select *,row_number() over(partition by id order by date ) 'rank'
from orde
where orders is not NULL
) a
) b
group by b.date_sub,id
having count(*) >= 3
;

查看數(shù)據(jù)：