Mysql-练习题

思路：
    获取所有有生物课程的人（学号，成绩） - 临时表
    获取所有有物理课程的人（学号，成绩） - 临时表
    根据【学号】连接两个临时表：
        学号  物理成绩   生物成绩
 
    然后再进行筛选

SELECT a_name,h_name,h_number FROM 
(select (student.student_name)as a_name, (class.class_name)as w_name, (score.number)as w_number from score
left join class on class.class_nid = class_sid
left join student on student_sid = student.student_nid
where class_name = '温泉') as a
LEFT JOIN
(select (student.student_name)as b_name, (class.class_name)as h_name, (score.number)as h_number  from score
left join class on class.class_nid = class_sid
left join student on student_sid = student.student_nid
where class_name = '化学') as b
on a_name = b_name
WHERE h_number > w_number 

SELECT a_name,avg(w_number) FROM 
(select (student.student_name)as a_name, (class.class_name)as w_name, (score.number)as w_number from score
left join class on class.class_nid = class_sid
left join student on student_sid = student.student_nid
) as a

GROUP BY a_name having avg(w_number) > 60

分组后对聚合函数进行操作 用having

SELECT * FROM 
(select (student.student_name)as a_name, (class.class_name)as w_name, (score.number)as w_number,(teacher.teacher_name)as t_name from score
left join class on class.class_nid = class_sid
left join student on student_sid = student.student_nid
left join teacher on class_teacher = teacher.teacher_nid
) as a
where t_name = '波多'

思路：
   先找到上过老师可的学生
   在学生表里面 去掉有过的学生 
    distinct 不同的  这里是去掉上过老师课重名的学生

select * from student
where student_name not in 
(select distinct student_id from score
LEFT JOIN course on course_id = course.course_name 
WHERE teacher_id = '加藤')

思路：
   先找到所有学过生物和温泉的所有学生，
   然后通过分组判断来显示
select * from 
(select student_id  from score
where course_id = '温泉' or course_id = '生物' ) as wq
GROUP BY student_id having count(student_id) > 1

思路:
   找到所有成绩 按照学号分组，判断是等于总科目的个数 

select * from 
(select student_id from score
group by student_id
having COUNT(student_id) < (select count(course_name) from course)
) as bb 

思路：
  先找到瞎子同学上的课有那些，
    然后链接所有上过瞎子课程的同学， 最后去除瞎子 课程总数少的 同学；

select student_id from 
(select (student_id) as ssd,(course_id)as sd from score
WHERE student_id = '瞎子'
) as sy
left join 
(SELECT * from score) as bb
on sd = bb.course_id
where student_id != '瞎子'
GROUP BY student_id HAVING count(student_id) > 1