-- 多对多的秘密：项目中，往往把关系分拆出一张单独表

--维表（学生维度）

create table t_stu(

id number primary key,

user_name varchar2(10)

);

insert into t_stu values(1,'小军');

insert into t_stu values(2,'小黄');

commit

select * from t_stu

drop table t_ke

--维表（课程维度）

create table t_ke(

id number primary key,

ke_name varchar2(100)

);

insert into t_ke values(1,'高等数学');

insert into t_ke values(2,'离散数学');

insert into t_ke values(3,'牛顿法');

insert into t_ke values(4,'大学英语');

insert into t_ke values(5,'马克思主义');

insert into t_ke values(6,'江泽明思想');

insert into t_ke values(7,'邓小平理论');

commit

select * from t_ke

-- 事实表

CREATE TABLE t_stu_ke(

id NUMBER PRIMARY KEY,

user_id NUMBER(5),

ke_id NUMBER(5)

);

delete from t_stu_ke

insert into t_stu_ke values(1,1,1);

insert into t_stu_ke values(2,1,2);

insert into t_stu_ke values(3,1,6);

commit;

insert into t_stu_ke values(4,2,4);

insert into t_stu_ke values(5,2,6);

commit;

select sk.id,s.user_name,k.ke_name from t_stu_ke sk

left join t_stu s on sk.user_id =s.id

left join t_ke k on sk.ke_id=k.id

insert into t_stu_ke values(6,3,3);

insert into t_stu_ke values(7,2,8);

commit

select

sk.id,

nvl(s.user_name,'未知学生:'||sk.user_id) 学生,

nvl(k.ke_name,'未知课程:'||sk.ke_id) 选课

from t_stu_ke sk

left join t_stu s on sk.user_id =s.id

left join t_ke k on sk.ke_id=k.id

-- 站在不通的维度看世界  

group by

--1.查询有多少科

select * from t_ke

select count(1) from t_ke

--2.查询同学选了多少课

select count（1） from

(select distinct ke_id from t_stu_ke) tt

--3.查选一下每个同学各选多少课

-- group by 有5个聚合函数count

select user_id,count(1) from t_stu_ke group by user_id

--4.查选一下每个同学各选多少课(条件是这个课程必须存在)

select user_id,count(1) from t_stu_ke

where ke_id in (select id from t_ke)

group by user_id

--5.查选一下每门课程各有多少学生选

select * from t_ke

--错误示范：

select ke_id,count(1) from t_stu_ke group by ke_id

--正确：

select ke_name,count(ke_id) from t_ke

left join t_stu_ke

on t_ke.id=t_stu_ke.ke_id

group by ke_name

--6.查选一下每门课程各有多少学生选

-- 条件：【显示人数】1人以上的课程

-- 思路：【显示人数】-->>集合数，不能用where，改用having

select ke_name,count(ke_id) r from t_ke

left join t_stu_ke

on t_ke.id=t_stu_ke.ke_id

group by ke_name

having count(ke_id)>=1

order by r desc

-- 7 查询出现毛的课程出来

select * from t_ke where ke_name like '%毛%'

-- 8 最多人选的课程是什么？

-- 思路1：

select * from t_ke where id=

(select ke_id from (

select ke_id,count(1) tt from t_stu_ke

group by ke_id order by tt desc

) where rownum=1)

--思路2(不推荐)

select * from t_ke where id=(

select ke_id from

(select ke_id,count(1) tt from t_stu_ke group by ke_id) where tt =

(select max(tt) from

(select ke_id,count(1) tt from t_stu_kegroup by ke_id) tt)

);