mysql练习（待补充）

2、查询‘生物’课程比‘物理’课程成绩高的所有学生的学号

思路：

获取所有生物课程的人（学号，成绩）-临时表

获取所有物理课程的人（学号，成绩）-临时表

根据学号连接两个临时表：

学号 生物成绩 物理成绩

SELECT A.student_id,sw,wl from
(SELECT student_id,num as sw from score LEFT JOIN course on score.course_id=course.cid where course.cname='生物')as A
LEFT JOIN
(SELECT student_id,num as wl from score LEFT JOIN course on score.course_id=course.cid where course.cname='物理')as B
on A.student_id =B.student_id where sw > if (ISNULL(wl),0,wl);

自己写了一段

SELECT A.student_id,sw,wl from
(SELECT student_id,num as sw from score where course_id=(SELECT cid from course WHERE cname='生物')) as A
LEFT JOIN
(SELECT student_id,num as wl from score where course_id=(SELECT cid from course WHERE cname='物理')) as B
on A.student_id=B.student_id where sw > if (ISNULL(wl),0,wl)

3、查询平均成绩大于60分的同学的学号和平均成绩

思路：

根据学生分组，使用avg获取平均值，通过having对avg进行筛选

select student_id,avg(num) from score group by student_id having avg(num) > 60

4、查询所有同学的学号、姓名、选课数、总成绩

select score.student_id,student.sname,count(score.student_id),sum(score.num)
from 
score left join student on score.student_id=student.sid group by score.student_id

5、查询姓‘李’的老师的个数

select count(tid) from teacher where tname like '李%'

6、查询没学过‘叶平’老师课的同学的学号、姓名

思路：

先查询‘李平老师’老师教的所有课的ID

然后获取选了李平老师课的学生ID

最后从学生表中筛选

select sid,sname from student where sid not in (
         select DISTINCT student_id from score where score.course_id in (
            select cid from course left join teacher on course.teacher_id = teacher.tid where tname = '李平老师'
        )
     )

7、查询学过‘001’并且也学过编号‘002’课程的同学的学号、姓名

思路：

先查到既选择001又选择002课程的所有同学

根据学生进行分组，如果学生数量等于2表示，两门均已选择

select student_id,sname FROM
(SELECT student_id,course_id from score where course_id =1 or course_id =2)as B
LEFT JOIN
student on B.student_id=student.sid GROUP BY student_id having count(student_id)>1

8、查询学过‘叶平’老师所教的所有课的同学的学号、姓名

select sid,sname from student where sid in (
         select DISTINCT student_id from score where score.course_id in (
            select cid from course left join teacher on course.teacher_id = teacher.tid where tname = '李平老师'
        )
     )

9、查询课程编号‘002’的成绩比课程编号‘001’课程低的所有同学的学号、姓名

同第一题，把大于号改成小于号

SELECT A.student_id,sw,wl from
(SELECT student_id,num as sw from score LEFT JOIN course on score.course_id=course.cid where course.cname='生物')as A
LEFT JOIN
(SELECT student_id,num as wl from score LEFT JOIN course on score.course_id=course.cid where course.cname='物理')as B
on A.student_id =B.student_id where sw < if (ISNULL(wl),0,wl)

我写了个更简单的

SELECT A.student_id,one,two from
(SELECT student_id,num as one FROM score where course_id=1)as A
LEFT JOIN
(SELECT student_id,num as two FROM score where course_id=2)as B
on A.student_id =B.student_id where one < if (ISNULL(two),0,two);

10、查询有课程成绩小于60分的同学的学号、姓名

SELECT sid,sname from student where sid in (SELECT DISTINCT student_id from score where num < 60 )

11、查询没学全所有课的同学的学号、姓名

思路：

在分数表中根据学生进行分组，获取每一个学生选课数量

如果数量==总课程数量，表示已经选择了所有课程

select student_id,sname from score left join student on score.student_id =student.sid group by student_id having count(course_id) =(select count(1) from course)

12、查询至少有一门课与学号为‘001’的同学所学相同的同学的学号和姓名（没有理解）

思路：

获取001同学选择的所有课程

获取课程在其中的所有人以及所有课程

根据学生筛选，获取所有学生信息

再与学生表连接，获取姓名

SELECT student_id,sname,count(course_id)
from score LEFT JOIN student on score.student_id=student.sid
where student_id!=1 and course_id in (SELECT course_id from score where student_id=1)GROUP BY student_id

13、查询至少学过学号为‘001’同学的所有课的其他同学学号和姓名（没有理解）

思路：

先找到和001的学过的所有人

然后个数=001所有学科   ==》》其他人可能选择的更多

select student_id,sname, count(course_id)
        from score left join student on score.student_id = student.sid
        where student_id != 1 and course_id in (select course_id from score where student_id = 1) group by student_id having count(course_id) ＝ (select count(course_id) from score where student_id = 1)

14、查询和“002”号的同学学习的课程完全相同的其他同学学号和姓名；

思路：      

个数相同

002学过的也学过

select student_id,sname from score left join student on score.student_id = student.sid where student_id in (
            select student_id from score  where student_id != 1 group by student_id HAVING count(course_id) = (select count(1) from score where student_id = 1)
        ) and course_id in (select course_id from score where student_id = 1) group by student_id HAVING count(course_id) = (select count(1) from score where student_id = 1)

15、删除学习‘李平’老师课的score表记录

delete from score where course_id in(select cid from course left join teacher on course.teacher_id = teacher.tid where teacher.tname = '李平老师')

不用连表写了一次

DELETE from score where course_id in (SELECT cid from course where teacher_id=(SELECT tid from teacher WHERE tname='李平老师'))

16、向sc表中插入一些记录，这些记录要求符合以下条件:没有上过编号‘002’课程的同学学号；插入'002'号课程的平均成绩

思路：

由于insert支持

insert into tb1(xx,xx) select x1,x2 from tb2;

所以，获取所有没上过002课的所有人，获取002的平均成绩

insert into score(student_id,course_id,num)
SELECT sid,2,(SELECT avg(num) from score where course_id=2)
from student WHERE sid not in (SELECT student_id from score where course_id=2)

17、按平均成绩从低到高，显示所有学生的‘生物’、‘物理’、‘体育’三门的课程成绩，按如下形式显示：学生ID,生物，物理，体育，有效课程数，有效平均分(没有理解）

select sc.student_id,
        (select num from score left join course on score.course_id = course.cid where course.cname = "生物" and score.student_id=sc.student_id) as sy,
        (select num from score left join course on score.course_id = course.cid where course.cname = "物理" and score.student_id=sc.student_id) as wl,
        (select num from score left join course on score.course_id = course.cid where course.cname = "体育" and score.student_id=sc.student_id) as ty,
        count(sc.course_id),
        avg(sc.num)
    from score as sc
    group by student_id desc

18、查询各科成绩最高和最低分：以如下形式显示：课程ID,最高分，最低分

select course_id,MAX(num) as max_num,min(num) as min_num from score GROUP BY course_id;

19、按各科平均成绩从低到高和及格率的百分数从高到低顺序(没有理解）

思路：

case when .. then

select course_id, avg(num) as avgnum,sum(case when score.num > 60 then 1 else 0 END)/count(1)*100 as percent from score group by course_id order by avgnum asc,percent desc;

20、课程平均分从高到低显示(现实任课老师)没有理解并且sql运行失败

select avg(if(isnull(score.num),0,score.num)),teacher.tname from course
    left join score on course.cid = score.course_id
    left join teacher on course.teacher_id = teacher.tid
    group by score.course_id