接着训练一

I、sql训练

21、查询不同老师所教不同课程平均分从高到低显示 

select
    t3.t_id,
    t3.t_name,
    avg(t1.s_score) as avg_score
from
    score t1 
    left join course t2 on t1.c_id = t2.c_id 
    left join teacher t3 on t2.t_id = t3.t_id
group by
    t3.t_id,
    t3.t_name
order by
    avg_score desc;

22、查询所有课程的成绩第2名到第3名的学生信息及该课程成绩

这里学习到rank（）函数

select * from student t1 join
(select s_id,s_score,c_id, (rank() over(partition by c_id order by s_score desc)) as 排名
from score ) as t2 on t1.s_id =t2.s_id

23、

select t2.c_id,t2.c_name,
(sum(case when s_score between 85 and 100 then 1 else 0 end)/count(s_score))*100 a,
(sum(case when s_score <85 and s_score>70 then 1 else 0 end)/count(s_score))*100 b,
(sum(case when s_score between 60 and 70 then 1 else 0 end)/count(s_score))*100 c,
(sum(case when s_score between 0 and 60 then 1 else 0 end)/count(s_score))*100 d
from score t1 join course t2 on t1.c_id=t2.c_id
group by t1.c_id

24、

select avg(s_score), rank() over(order by avg(s_score)) from score group by s_id 

25、

select * 
from
(select c_id,s_id,dense_rank() over(partition by c_id order by s_score) as ran from score) t1
where t1.ran<4

26、

select c_id,count(s_id) from score group by c_id

27、

select s_id,s_name from student where s_id in
(select s_id from score group by s_id having  count(s_score) = 2)

28、

select s_sex,count(s_id) from student group by s_sex;

29、

select * from student where s_name like '%风%'

30、

select s_name,count(s_id) as count_num from student group by s_name having count(s_id)>1

31、

select * from student where s_birth like '1990%'

32、

连续两个条件排序，只需要加个条件即可

select avg(s_score) from score group by c_id order by avg(s_score) desc,c_id asc;

33、

select t1.s_id,t1.s_name,avg(t2.s_score) as avg_score from student t1 join score t2 on t1.s_id=t2.s_id group by t1.s_id having avg_score >=85

34、

select t1.s_name,t2.s_score from student t1 join score t2 on t1.s_id =t2.s_id join course t3 on t2.c_id =t3.c_id where t3.c_name='数学' and t2.s_score <60