mysql习题

　　如图表创建数据库。

create table class(
                    cid int auto_increment primary key,
                    caption char(20)
                )engine=innodb default charset=utf8;
create table student(
                    sid int auto_increment primary key,
                    sname char(10),
                    gender char(10),
                    class_id int not null,
                    CONSTRAINT fk_cla_cid FOREIGN key (class_id) REFERENCES class(cid)
                )engine=innodb default charset=utf8;
                select * from student,class where student.class_id = class.cid;
create table teacher(
                    tid int auto_increment primary key,
                    tname char(20)
                )engine=innodb default charset=utf8;
create table course(
                    cid int auto_increment primary key,
                    cname char(10),
                    teach_id int not null,
                    CONSTRAINT fk_tea_tid FOREIGN key (teach_id) REFERENCES teacher(tid)
                )engine=innodb default charset=utf8;
select * from course,teacher where course.teach_id = teacher.tid;
create table score(
                    sid int auto_increment primary key,
                    student_id int not null,
                    course_id int not null,
                    number int,
                    unique uq_stu_course (student_id,course_id),
                    CONSTRAINT fk_score_stu FOREIGN key (student_id) REFERENCES student(sid),
                    CONSTRAINT fk_score_course FOREIGN key (course_id) REFERENCES course(cid)
                )engine=innodb default charset=utf8;

数据库创建

　　1.查询所有成绩大于60分的人；

select * from score where number >60;
+-----+------------+-----------+--------+
| sid | student_id | course_id | number |
+-----+------------+-----------+--------+
|   3 |          2 |         2 |    100 |
+-----+------------+-----------+--------+

成绩大于60

　　2.查询每个老师有几个学生；

mysql> select teach_id,count(cname) from course group by teach_id;
+----------+--------------+
| teach_id | count(cname) |
+----------+--------------+
|        1 |            2 |
|        2 |            1 |
+----------+--------------+

几人选修

　　3.查询老师名字与课程的关联（连表）；

mysql> select * from course left join teacher on course.teach_id = teacher.tid;
+-----+--------+----------+------+--------+
| cid | cname  | teach_id | tid  | tname  |
+-----+--------+----------+------+--------+
|   1 | 生物   |        1 |    1 | 波多   |
|   2 | 体育   |        1 |    1 | 波多   |
|   3 | 物理   |        2 |    2 | 苍空   |
+-----+--------+----------+------+--------+

老师对应课程

　　4.查询男女生人数；

mysql> select gender,count(sid) from student group by gender;
+--------+------------+
| gender | count(sid) |
+--------+------------+
| 女     |          2 |
| 男     |          1 |
+--------+------------+

查询男女生人数

　　5.查看平均成绩大于50的人的信息；

mysql> select grade.student_id,student.sname,grade.av from (select student_id,avg(number)as av from score group by student_id having avg(number) > 50) as grade
    -> left join student on grade.student_id = student.sid;
+------------+--------+----------+
| student_id | sname  | av       |
+------------+--------+----------+
|          1 | 钢弹   |  59.5000 |
|          2 | 铁锤   | 100.0000 |
+------------+--------+----------+

平均分+连表

　　6.查询所有同学的学号、姓名、选课数、总成绩；

mysql> select student.sid,student.sname,count(1),sum(number) from score left join student on student.sid = score.student_id group by score.student_id;
+------+--------+----------+-------------+
| sid  | sname  | count(1) | sum(number) |
+------+--------+----------+-------------+
|    1 | 钢弹   |        2 |         119 |
|    2 | 铁锤   |        1 |         100 |
+------+--------+----------+-------------+

总成绩

　　7.查询没学过“波多”老师课的同学的学号、姓名；

mysql> select student.sid,student.sname from (select score.student_id as iid from score where course_id not in (select course.cid from course left join teacher on course.teach_id = teacher.tid where teacher.tname = '苍空') group by score.student_id)as i left join student on i.iid = student.sid;
+------+--------+
| sid  | sname  |
+------+--------+
|    1 | 钢弹   |
|    2 | 铁锤   |
+------+--------+
#select course.cid from course left join teacher on course.teach_id = teacher.tid where teacher.tname = '苍空'，这个老师的所有课的id，
接着是所有没选这个老师课的学生id的分组列表，最后和学生表连表得到结果

没选的人

mysql> select course.cid from course left join teacher on course.teach_id = teacher.tid where teacher.tname = '苍空';
+-----+
| cid |
+-----+
|   3 |
+-----+
这是苍空老师教的课；
mysql> select score.student_id from score where course_id in (select course.cid from course left join teacher on course.teach_id = teacher.tid where teacher.tname = '苍空') group by score.student_id;
Empty set (0.00 sec)这是从成绩表查询谁上了苍空老师的课，

select student.sid,student.sname from student where sid not in
(select score.student_id from score where course_id in (select course.cid from course left join teacher on course.teach_id = teacher.tid where teacher.tname = '苍空') group by score.student_id);
在学生表中排出这几个人就是没选他的课的人了。

选课

　　8.查询“生物”课程比“体育”课程成绩高的所有学生的学号；

mysql> select A.student_id from
    -> (select * from score left join course on score.course_id = course.cid where course.cname='生物') as A
    -> inner join
    -> (select * from score left join course on score.course_id = course.cid where course.cname='体育') as B
    -> on A.student_id = B.student_id where A.number>B.number;
+------------+
| student_id |
+------------+
|          1 |
+------------+

　　9.查询学过“1”并且也学过编号“2”课程的同学的学号、姓名；

mysql> select score.student_id,student.sname from score left join student
    -> on score.student_id=student.sid
    ->  where score.course_id =1 or score.course_id =2 group by student_id having count(course_id)>1;
+------------+--------+
| student_id | sname  |
+------------+--------+
|          1 | 钢弹   |
+------------+--------+

学过两门课

　　10.查询学过“波多”老师所教的所有课的同学的学号；

select cid from course left join teacher on course.teach_id = teacher.tid where teacher.tname='波多';
波多老师上课的id
mysql> select student_id from score where course_id in
    -> (select cid from course left join teacher on course.teach_id = teacher.tid where teacher.tname='波多')
    -> group by student_id having count(course_id)=
    -> (select count(cid) from course left join teacher on course.teach_id = teacher.tid where teacher.tname='波多');
+------------+
| student_id |
+------------+
|          1 |
+------------+
成绩表里上过他的课的人都列出来，然后进行分组，学过他的课与他带的课一样多就是要查询的学生id

老师所有的课

　　11.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；

mysql> select A.student_id from
    -> (select * from score where course_id=1) as A
    -> inner join
    -> (select * from score where course_id=2) as B
    -> on A.student_id = B.student_id where A.number>B.number;
+------------+
| student_id |
+------------+
|          1 |
+------------+

和1差不多

　　12.查询有课程成绩小于61分的同学的学号、姓名；

mysql> select student.sid,student.sname from student where student.sid in
    -> (select student_id from score where number<61 group by student_id);
+-----+--------+
| sid | sname  |
+-----+--------+
|   1 | 钢弹   |
+-----+--------+

低于61分

　　13.查询没有学全所有课的同学的学号、姓名；

mysql> select student_id,student.sname from score left join student on score.student_id = student.sid
    -> group by student_id having count(1)<(select count(1) from course);
+------------+--------+
| student_id | sname  |
+------------+--------+
|          1 | 钢弹   |
|          2 | 铁锤   |
+------------+--------+

没学全的人

　　14.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；

mysql> select score.student_id,student.sname 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;
+------------+--------+
| student_id | sname  |
+------------+--------+
|          2 | 铁锤   |
+------------+--------+

至少一门课和他一样

　　15.查询至少学过学号为“001”同学所有课的其他同学学号和姓名；

mysql> select score.student_id,student.sname from score
    -> left join student on score.student_id = student.sid
    ->  where student_id !=2 and course_id in (select course_id from score where student_id = 2)
    -> group by student_id
    -> having count(course_id)=(select count(course_id) from score where student_id = 2);
+------------+--------+
| student_id | sname  |
+------------+--------+
|          1 | 钢弹   |
+------------+--------+

至少都学过

　　16.查询和“002”号的同学学习的课程完全相同的其他同学学号;

mysql> select student_id from score where student_id in (
    -> select student_id from score where student_id !=2 group by student_id having count(1) =(select count(1) from score where student_id = 2))
    -> and course_id in (select course_id from score where student_id = 2) group by student_id having count(1)=(select count(1) from score where student_id = 2);
Empty set (0.00 sec)

and之前找到和2上课一样所有的id，and之后这个人上的所有课的id也要和2同学一样

select count(1) from score where student_id = 2;#2同学上的课个数

select student_id from score where student_id !=2 group by student_id having count(1) =(select count(1) from score where student_id = 2);
#和2上课数量一样多的同学

完全一样

　　17.删除学习“波多”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.name = '波多');
取出老师上的所有课的id再记录表中删除

删除表

　　18.向score表中插入一些记录，这些记录要求符合以下条件：①没有上过编号“002”课程的同学学号；②插入“002”号课程的平均成绩；

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);

没上过课程的信息

　　19.按平均成绩从低到高 显示所有学生的“语文”、“数学”、“英语”三门的课程成绩;

mysql> select sc.student_id,
    -> (select number from score left join course on score.course_id = course.cid where course.cname = "生物" and score.student_id=sc.student_id) as '生物',
    -> (select number from score left join course on score.course_id = course.cid where course.cname = "物理" and score.student_id=sc.student_id) as '物理',
    -> (select number from score left join course on score.course_id = course.cid where course.cname = "体育" and score.student_id=sc.student_id) as '体育',
    -> avg(sc.number)
    -> from score as sc
    -> group by student_id asc;
+------------+--------+--------+--------+----------------+
| student_id | 生物   | 物理   | 体育   | avg(sc.number) |
+------------+--------+--------+--------+----------------+
|          1 |     60 |   NULL |     59 |        59.5000 |
|          2 |   NULL |   NULL |    100 |       100.0000 |
+------------+--------+--------+--------+----------------+

数据库的循环操作

　　20.查询各科成绩最高和最低的分；

mysql> select student_id,max(number),min(number) from score group by student_id;
+------------+-------------+-------------+
| student_id | max(number) | min(number) |
+------------+-------------+-------------+
|          1 |          60 |          59 |
|          2 |         100 |         100 |
+------------+-------------+-------------+

max，min

　　在此基础上如果我想要让低于60分的都为0显示，需要用到sql的判断语句；

mysql> select student_id,max(number),min(number),min(number)+1,case when min(number)<60 then 0 else min(number) end as  成绩 from score group by student_id;
+------------+-------------+-------------+---------------+--------+
| student_id | max(number) | min(number) | min(number)+1 | 成绩   |
+------------+-------------+-------------+---------------+--------+
|          1 |          60 |          59 |            60 |      0 |
|          2 |         100 |         100 |           101 |    100 |
+------------+-------------+-------------+---------------+--------+

case when。。。

　　21.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；

mysql> select course_id, avg(number) as avgnum,sum(case when score.number > 60 then 1 else 0 END)/count(1)*100 as percent from score group by course_id order by avgnum asc,percent desc;
+-----------+---------+---------+
| course_id | avgnum  | percent |
+-----------+---------+---------+
|         1 | 60.0000 |  0.0000 |
|         2 | 79.5000 | 50.0000 |
+-----------+---------+---------+
#大于60分标记成1否则为0，count(1)或者用sum（1）这是单独摆成一列为这科考试的总人数，相当于但都有一列全都为1，然后除左边的合格人数1除以右边总人数就是及格率了。

及格率

　　22.课程平均分从高到低显示；

mysql> select avg(if(isnull(score.number),0,score.number)),teacher.tname from score
    ->     left join course on  score.course_id = course.cid
    ->     left join teacher on teacher.tid = course.teach_id
    ->     group by score.course_id;
+----------------------------------------------+--------+
| avg(if(isnull(score.number),0,score.number)) | tname  |
+----------------------------------------------+--------+
|                                      60.0000 | 波多   |
|                                      79.5000 | 波多   |
+----------------------------------------------+--------+
if（isnull（A）B,C）就是sql的三目运算A为空则结果为B，否则为C

　　23.查询各科成绩前三名的记录;

mysql> select * from
    -> (
    -> select
    -> student_id,
    -> course_id,
    -> number,
    -> 1,
    -> (select number from score as s2 where s2.course_id = s1.course_id group by s2.number order by s2.number desc limit 0,1) as first_num,
    -> (select number from score as s2 where s2.course_id = s1.course_id group by s2.number order by s2.number desc limit 1,1) as third_num
    -> from
    -> score as s1
    -> ) as T
    -> where T.number > T.third_num;
+------------+-----------+--------+---+-----------+-----------+
| student_id | course_id | number | 1 | first_num | third_num |
+------------+-----------+--------+---+-----------+-----------+
|          2 |         2 |    100 | 1 |       100 |        59 |
+------------+-----------+--------+---+-----------+-----------+
select number from score as s2 where s2.course_id = s1.course_id group by s2.number order by s2.number desc limit 0,1#第一名并列的成绩并排序

各科前三

　　24.查询每门课程被选修的学生数;

select course_id, count(1) from score group by course_id;
+-----------+----------+
| course_id | count(1) |
+-----------+----------+
|         1 |        1 |
|         2 |        2 |
+-----------+----------+

课程人数

　　查询出只选修了一门课程的全部学生的学号和姓名;

mysql> select student.sid, student.sname,count(1) from score
    -> left join student on score.student_id  = student.sid
    -> group by student_id having count(1) = 1;
+------+--------+----------+
| sid  | sname  | count(1) |
+------+--------+----------+
|    2 | 铁锤   |        1 |
+------+--------+----------+

　　25.查询男生、女生的人数；

mysql> select gender,count(1) from student group by student.gender;
+--------+----------+
| gender | count(1) |
+--------+----------+
| 女     |        2 |
| 男     |        1 |
+--------+----------+

男女人数

　　26.查询姓“钢”的学生名单；

mysql> select sname from student where sname like '钢%';
+--------+
| sname  |
+--------+
| 钢弹   |
+--------+

　　27.查询同名同姓学生名单，并统计同名人数；

mysql> select sname,count(1) from student group by sname having count(1)>1;
Empty set (0.00 sec)

　　28.查询每门课程的平均成绩，结果按平均成绩升序排列，平均成绩相同时，按课程号降序排列;

select course_id,avg(if(isnull(number), 0 ,number)) as avg from score group by course_id order by avg asc,course_id desc;

　　29.查询平均成绩大于85的所有学生的学号、姓名和平均成绩;

mysql> select student_id,sname,avg(if(isnull(number), 0 ,number)) as avg from score left join student on score.student_id = student.sid group by student_id having avg >85;
+------------+--------+----------+
| student_id | sname  | avg      |
+------------+--------+----------+
|          2 | 铁锤   | 100.0000 |
+------------+--------+----------+
先列出所有的平均值，再判断取出

avg大于85

　　30.查询课程名称为“体育”，且分数低于60的学生姓名和分数;

mysql> select student.sname,score.number from score
    -> left join course on score.course_id = course.cid
    -> left join student on score.student_id = student.sid
    -> where score.number < 60 and course.cname = '体育';
+--------+--------+
| sname  | number |
+--------+--------+
| 钢弹   |     59 |
+--------+--------+

固定课程的成绩不及格

　　31.查询选修“波多”老师所授课程的学生中，成绩最高的学生姓名及其成绩；

select course.cid from course left join teacher on course.teach_id = teacher.tid where tname='波多';
+-----+
| cid |
+-----+
|   1 |
|   2 |
+-----+
先找出波多老师教授的课程
mysql> select sname,number from score
    ->     left join student on score.student_id = student.sid
    ->     where score.course_id in (select course.cid from course left join teacher on course.teach_id = teacher.tid where tname='波多') order by number desc limit 1;
+--------+--------+
| sname  | number |
+--------+--------+
| 铁锤   |    100 |
+--------+--------+

　　32.查询不同课程但成绩相同的学生的学号、课程号、学生成绩；

select DISTINCT s1.course_id,s2.course_id,s1.number,s2.number from score as s1, score as s2 where s1.number = s2.number and s1.course_id != s2.course_id;