【SQL练习题】找出右图中，id的申请和受理日期之间超过了10个工作日的id

找出右图中，id的申请和受理日期之间超过了10个工作日的id

左表如下:

右表如下：

—1.建a表

create table a(
t date,
status varchar2(10)
);

—2.a表中插入数据

insert into a values(date'2020-6-1','Y');
insert into a values(date'2020-6-2','Y');
insert into a values(date'2020-6-3','Y');
insert into a values(date'2020-6-4','Y');
insert into a values(date'2020-6-5','Y');
insert into a values(date'2020-6-6','N');
insert into a values(date'2020-6-7','N');
insert into a values(date'2020-6-8','Y');
insert into a values(date'2020-6-9','Y');
insert into a values(date'2020-6-10','Y');
insert into a values(date'2020-6-11','Y');
insert into a values(date'2020-6-12','Y');
insert into a values(date'2020-6-13','N');
insert into a values(date'2020-6-14','N');
insert into a values(date'2020-6-15','Y');
insert into a values(date'2020-6-16','Y');
insert into a values(date'2020-6-17','Y');
insert into a values(date'2020-6-18','Y');
insert into a values(date'2020-6-19','Y');
insert into a values(date'2020-6-20','N');
insert into a values(date'2020-6-21','N');
insert into a values(date'2020-6-22','Y');
insert into a values(date'2020-6-23','Y');
insert into a values(date'2020-6-24','Y');
insert into a values(date'2020-6-25','N');
insert into a values(date'2020-6-26','N');
insert into a values(date'2020-6-27','N');
insert into a values(date'2020-6-28','Y');
insert into a values(date'2020-6-29','Y');
insert into a values(date'2020-6-30','Y');
commit;

—3.建b表

create table b(
id varchar2(10),
sdate date,
ldate date
);

—4.b表中插入数据

insert into b values('S001',date'2020-6-2',date'2020-6-8');
insert into b values('S002',date'2020-6-3',date'2020-6-26');
insert into b values('S003',date'2020-6-5',date'2020-6-15');
insert into b values('S004',date'2020-6-15',date'2020-6-30');
insert into b values('S005',date'2020-6-18',date'2020-6-27');
commit;

—5.查询语句

select id
  from a
  join b
    on a.t between b.sdate and b.ldate
   and status = 'Y'
 group by id
having count(1) > 10;