經典SQL問題：Top 10%

學生表：

create table hy_student(
   id number(4,0) primary key,
   name nvarchar2(20) not null,
   score number(3,0) not null)

充值：

insert into hy_student
select rownum,dbms_random.string('*',dbms_random.value(1,20)),dbms_random.value(0,100)
from dual
connect by level<201
order by dbms_random.random

第一步把學生按成績逆序排列：

select * from hy_student order by score desc

第二步給加偽列：

select rownum as rn,a.* from (select * from hy_student order by score desc) a

最后就可以檢出前10%的精英了：

select b.* from (select rownum as rn,a.* from (select * from hy_student order by score desc) a) b where b.rn<= (select count(*) from hy_student)/10

結果：

--2020-04-02--

以上用到的全部SQL：

create table hy_student(
   id number(4,0) primary key,
   name nvarchar2(20) not null,
   score number(3,0) not null)
   
insert into hy_student
select rownum,dbms_random.string('*',dbms_random.value(1,20)),dbms_random.value(0,100)
from dual
connect by level<201
order by dbms_random.random

commit;

select * from hy_student order by score desc

select rownum as rn,a.* from (select * from hy_student order by score desc) a

select b.* from (select rownum as rn,a.* from (select * from hy_student order by score desc) a) b where b.rn<= (select count(*) from hy_student)/10