MySQL50-8-第26-30题

select
	c.c_id
	,c.c_name
	,count(s.s_id)
from Course c
join Score s
on c.c_id = s.c_id
group by c.c_id;

select
	s.s_id
	,s.s_name
from Student s
join Score sc
on s.s_id = sc.s_id
group by 1,2
having count(sc.c_id) = 2;

--  自己的方法
select
	count(case when s_sex='男' then 1 end) as '男'
	,count(case when s_sex='女' then 1 end) as '女'
from Student;

-- 参考方法
select
	s_sex
	,count(s_sex) as `人数`
from Student
group by s_sex;

-- 模糊匹配：我们在两边都加上了%，考虑的是姓或者名字含有风，虽然风姓很少见
select * from Student where s_name like "%风%";

select
	a.s_name
	,a.s_sex
	,count(*)
from Student a  -- 同一个表的自连接
join Student b
on a.s_id != b.s_id   -- 连接的时候不能是同一个人：学号保证，每个人的学号是唯一的，其他字段都可能重复
and a.s_sex = b.s_sex  -- 性别相同
and a.s_name = b.s_name -- 名字相同
group by 1,2;