[MySQL] SQL查询至少连续七天下单的用户

create table 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 )

select *,date-排名 rnk from (select *,row_number() over(partition by id order by date) 排名 from orde) a;

select id,rnk,count(*) ok from
(select *,date-排名 rnk from (select *,row_number() over(partition by id order by date) 排名 from orde) a) b group by id,rnk;

select id,rnk from
(select *,date-排名 rnk from (select *,row_number() over(partition by id order by date) 排名 from orde) a) b group by id,rnk having count(*)>=7;