sql 五十题 26-30

zzz大王2025-12-13 13:26

sql 复制代码

#26、查询每门课程被选修的学生数 
select
count(a.s_id) as count_course , b.c_name
from
score a
left join 
course b
on
a.c_id = b.c_id
group by a.c_id

书写怎么这么糟糕......

sql 复制代码

 #27、查询出只有两门课程的全部学生的学号和姓名
 select count(a.c_id),b.*
 from score a, student b
 where a.s_id = b.s_id
 group by a.s_id
 having count(a.c_id) = 2

sql 复制代码

#查询男生女生人数
select
count(s_sex),s_sex
from
student
group by s_sex

666count总是搞不懂应该count谁，用谁分类。

sql 复制代码

-- 29、查询名字中含有"风"字的学生信息
select
*
from
student
where s_name like '%风'

注意like 和 = 的区别

sql 复制代码

-- 30、查询同名同性学生名单，并统计同名人数
with t_table as 
	(select count(s_name) over (partition by s_id order by count(s_name)) as count_name,
		s_name,
		s_id,
		s_sex
	from student 
	group by s_id )
select s_name,s_sex,count(*)
from
t_table
where count_name > 1
group by s_id

明显你写的很复杂，看解析

sql 复制代码

select
	a.s_name,
    a.s_sex,
    count(*) 
from student a 
JOIN 
	student b 
on a.s_id !=b.s_id 
and a.s_name = b.s_name 
and a.s_sex = b.s_sex
GROUP BY a.s_name,a.s_sex

首先两个表连接做笛卡尔积，随后限定条件有【a.s_id != b.s_id】这个条件出来就是异响匹配，你写的时候没想到，然后是两个连接条件，再是【count（*）】是计算做出来的表格所有的行数，也就是只符合条件的总人数。