mysql-面试50题-2

 一、查询数据

学生表 Student

create table Student(SId varchar(10),Sname varchar(10),Sage datetime,Ssex varchar(10));

insert into Student values('01' , '赵雷' , '1990-01-01' , '男');

insert into Student values('02' , '钱电' , '1990-12-21' , '男');

insert into Student values('03' , '孙风' , '1990-12-20' , '男');

insert into Student values('04' , '李云' , '1990-12-06' , '男');

insert into Student values('05' , '周梅' , '1991-12-01' , '女');

insert into Student values('06' , '吴兰' , '1992-01-01' , '女');

insert into Student values('07' , '郑竹' , '1989-01-01' , '女');

insert into Student values('09' , '张三' , '2017-12-20' , '女');

insert into Student values('10' , '李四' , '2017-12-25' , '女');

insert into Student values('11' , '李四' , '2012-06-06' , '女');

insert into Student values('12' , '赵六' , '2013-06-13' , '女');

insert into Student values('13' , '孙七' , '2014-06-01' , '女');

科目表 Course

create table Course(CId varchar(10),Cname nvarchar(10),TId varchar(10));

insert into Course values('01' , '语文' , '02');

insert into Course values('02' , '数学' , '01');

insert into Course values('03' , '英语' , '03');

教师表 Teacher

create table Teacher(TId varchar(10),Tname varchar(10));

insert into Teacher values('01' , '张三');

insert into Teacher values('02' , '李四');

insert into Teacher values('03' , '王五');

成绩表 SC

create table SC(SId varchar(10),CId varchar(10),score decimal(18,1));

insert into SC values('01' , '01' , 80);

insert into SC values('01' , '02' , 90);

insert into SC values('01' , '03' , 99);

insert into SC values('02' , '01' , 70);

insert into SC values('02' , '02' , 60);

insert into SC values('02' , '03' , 80);

insert into SC values('03' , '01' , 80);

insert into SC values('03' , '02' , 80);

insert into SC values('03' , '03' , 80);

insert into SC values('04' , '01' , 50);

insert into SC values('04' , '02' , 30);

insert into SC values('04' , '03' , 20);

insert into SC values('05' , '01' , 76);

insert into SC values('05' , '02' , 87);

insert into SC values('06' , '01' , 31);

insert into SC values('06' , '03' , 34);

insert into SC values('07' , '02' , 89);

insert into SC values('07' , '03' , 98);

二、问题

11.查询两门及其以上不及格课程的同学的学号，姓名及其平均成绩

从SC表中选取score小于60的，并group by sid，having count 大于1，然后再在sc表中，查询平均成绩，最后与student表联合查询。

mysql> select student.SId, student.Sname,b.avg
    -> from student RIGHT JOIN
    -> (select sid, AVG(score) as avg from sc
    ->     where sid in (
    ->               select sid from sc
    ->               where score<60
    ->               GROUP BY sid
    ->               HAVING count(score)>1)
    ->     GROUP BY sid) b on student.sid=b.sid;

+------+--------+----------+
| SId  | Sname  | avg      |
+------+--------+----------+
| 04   | 李云   | 33.33333 |
| 06   | 吴兰   | 32.50000 |
+------+--------+----------+
2 rows in set (0.06 sec)

12.检索" 01 "课程分数小于 60，按分数降序排列的学生信息

这道题简单，就不多说了。

mysql> select student.*, sc.score from student, sc
    -> where student.sid = sc.sid
    -> and sc.score < 60
    -> and cid = "01"
    -> order by sc.score desc;

+------+--------+---------------------+------+-------+
| SId  | Sname  | Sage                | Ssex | score |
+------+--------+---------------------+------+-------+
| 04   | 李云   | 1990-12-06 00:00:00 | 男   |  50.0 |
| 06   | 吴兰   | 1992-01-01 00:00:00 | 女   |  31.0 |
+------+--------+---------------------+------+-------+
2 rows in set (0.00 sec)

13.按平均成绩从高到低显示所有学生的所有课程的成绩以及平均成绩

我的评价是，逻辑上没多大难度，写上也没有多少难度

mysql> select *  from sc
    -> left join (
    ->     select sid,avg(score) as avscore from sc
    ->     group by sid
    ->     )r
    -> on sc.sid = r.sid
    -> order by avscore desc;

+------+------+-------+------+----------+
| SId  | CId  | score | sid  | avscore  |
+------+------+-------+------+----------+
| 07   | 02   |  89.0 | 07   | 93.50000 |
| 07   | 03   |  98.0 | 07   | 93.50000 |
| 01   | 03   |  99.0 | 01   | 89.66667 |
| 01   | 02   |  90.0 | 01   | 89.66667 |
| 01   | 01   |  80.0 | 01   | 89.66667 |
| 05   | 02   |  87.0 | 05   | 81.50000 |
| 05   | 01   |  76.0 | 05   | 81.50000 |
| 03   | 01   |  80.0 | 03   | 80.00000 |
| 03   | 03   |  80.0 | 03   | 80.00000 |
| 03   | 02   |  80.0 | 03   | 80.00000 |
| 02   | 02   |  60.0 | 02   | 70.00000 |
| 02   | 01   |  70.0 | 02   | 70.00000 |
| 02   | 03   |  80.0 | 02   | 70.00000 |
| 04   | 01   |  50.0 | 04   | 33.33333 |
| 04   | 03   |  20.0 | 04   | 33.33333 |
| 04   | 02   |  30.0 | 04   | 33.33333 |
| 06   | 01   |  31.0 | 06   | 32.50000 |
| 06   | 03   |  34.0 | 06   | 32.50000 |
+------+------+-------+------+----------+
18 rows in set (0.00 sec)

14.查询各科成绩最高分、最低分和平均分：

以如下形式显示：课程 ID，课程 name，最高分，最低分，平均分，及格率，中等率，优良率，优秀率，及格为>=60，中等为：70-80，优良为：80-90，优秀为：>=90

要求输出课程号和选修人数，查询结果按人数降序排列，若人数相同，按课程号升序排列

这道题看起来要求很多，但从下面代码可以看出来，其实，这就是一个单表查询，其难点在于限制条件的书写。

mysql> select
    -> sc.CId ,
    -> max(sc.score)as 最高分,
    -> min(sc.score)as 最低分,
    -> AVG(sc.score)as 平均分,
    -> count(*)as 选修人数,
    -> sum(case when sc.score>=60 then 1 else 0 end )/count(*)as 及格率,
    -> sum(case when sc.score>=70 and sc.score<80 then 1 else 0 end )/count(*)as 中等率,
    -> sum(case when sc.score>=80 and sc.score<90 then 1 else 0 end )/count(*)as 优良率,
    -> sum(case when sc.score>=90 then 1 else 0 end )/count(*)as 优秀率
    -> from sc
    ->group by sc.CId
    -> order by count(*) desc, sc.CId asc;

+------+-----------+-----------+-----------+--------------+-----------+-----------+-----------+-----------+
| CId  | 最高分    | 最低分    | 平均分    | 选修人数     | 及格率    | 中等率    | 优良率    | 优秀率    |
+------+-----------+-----------+-----------+--------------+-----------+-----------+-----------+-----------+
| 01   |      80.0 |      31.0 |  64.50000 |            6 |    0.6667 |    0.3333 |    0.3333 |    0.0000 |
| 02   |      90.0 |      30.0 |  72.66667 |            6 |    0.8333 |    0.0000 |    0.5000 |    0.1667 |
| 03   |      99.0 |      20.0 |  68.50000 |            6 |    0.6667 |    0.0000 |    0.3333 |    0.3333 |
+------+-----------+-----------+-----------+--------------+-----------+-----------+-----------+-----------+
3 rows in set (0.00 sec)

15.按各科成绩进行排序，并显示排名， Score 重复时保留名次空缺

mysql>  select a.cid, a.sid, a.score, count(b.score)+1 as rank
    -> from sc as a
    -> left join sc as b
    -> on a.score     -> group by a.cid, a.sid,a.score
    -> order by a.cid, rank asc;

+------+------+-------+------+
| cid  | sid  | score | rank |
+------+------+-------+------+
| 01   | 01   |  80.0 |    1 |
| 01   | 03   |  80.0 |    1 |
| 01   | 05   |  76.0 |    3 |
| 01   | 02   |  70.0 |    4 |
| 01   | 04   |  50.0 |    5 |
| 01   | 06   |  31.0 |    6 |
| 02   | 01   |  90.0 |    1 |
| 02   | 07   |  89.0 |    2 |
| 02   | 05   |  87.0 |    3 |
| 02   | 03   |  80.0 |    4 |
| 02   | 02   |  60.0 |    5 |
| 02   | 04   |  30.0 |    6 |
| 03   | 01   |  99.0 |    1 |
| 03   | 07   |  98.0 |    2 |
| 03   | 03   |  80.0 |    3 |
| 03   | 02   |  80.0 |    3 |
| 03   | 06   |  34.0 |    5 |
| 03   | 04   |  20.0 |    6 |
+------+------+-------+------+
18 rows in set (0.00 sec)

16. 查询学生的总成绩，并进行排名，总分重复时不保留名次空缺

mysql> select q.sid, total, @crank := @crank +1 as rank from(
    -> select sc.sid, sum(sc.score) as total from sc
    -> group by sc.sid
    -> order by total desc)q;

+------+-------+------+
| sid  | total | rank |
+------+-------+------+
| 01   | 269.0 |    1 |
| 03   | 240.0 |    2 |
| 02   | 210.0 |    3 |
| 07   | 187.0 |    4 |
| 05   | 163.0 |    5 |
| 04   | 100.0 |    6 |
| 06   |  65.0 |    7 |
+------+-------+------+
7 rows in set (0.00 sec)

17.统计各科成绩各分数段人数：课程编号，课程名称，[100-85]，[85-70]，[70-60]，[60-0] 及所占百分比

mysql> select sc.cid,
    -> max(sc.score) 最高分,
    -> min(sc.score) 最低分,
    -> avg(sc.score) 平均分,
    -> count(*) 参选人数,
    -> sum(case when sc.score>=60 then 1 else 0 end)/count(*) "[60-0]",
    -> sum(case when sc.score>=60 and sc.score<70 then 1 else 0 end)/count(*) "[70-60]",
    -> sum(case when sc.score>=70 and sc.score<85 then 1 else 0 end)/count(*) "[85-70]",
    -> sum(case when sc.score>=85 then 1 else 0 end)/count(*) "[100-85]"
    -> from sc
    -> group by sc.cid
    -> order by sc.cid asc;

+------+-----------+-----------+-----------+--------------+--------+---------+---------+----------+
| cid  | 最高分    | 最低分    | 平均分    | 参选人数     | [60-0] | [70-60] | [85-70] | [100-85] |
+------+-----------+-----------+-----------+--------------+--------+---------+---------+----------+
| 01   |      80.0 |      31.0 |  64.50000 |            6 | 0.6667 |  0.0000 |  0.6667 |   0.0000 |
| 02   |      90.0 |      30.0 |  72.66667 |            6 | 0.8333 |  0.1667 |  0.1667 |   0.5000 |
| 03   |      99.0 |      20.0 |  68.50000 |            6 | 0.6667 |  0.0000 |  0.3333 |   0.3333 |
+------+-----------+-----------+-----------+--------------+--------+---------+---------+----------+
3 rows in set (0.00 sec)

18.查询各科成绩前三名的记录

mysql> select * from sc
    -> where (
    -> select count(*) from sc as a
    -> where sc.cid = a.cid and sc.score     -> )< 3
    -> order by cid asc, sc.score desc;

+------+------+-------+
| SId  | CId  | score |
+------+------+-------+
| 01   | 01   |  80.0 |
| 03   | 01   |  80.0 |
| 05   | 01   |  76.0 |
| 01   | 02   |  90.0 |
| 07   | 02   |  89.0 |
| 05   | 02   |  87.0 |
| 01   | 03   |  99.0 |
| 07   | 03   |  98.0 |
| 02   | 03   |  80.0 |
| 03   | 03   |  80.0 |
+------+------+-------+
10 rows in set (0.00 sec)

19.查询每门课程被选修的学生数

mysql> select cid, count(sid) from sc
    -> group by cid;

+------+------------+
| cid  | count(sid) |
+------+------------+
| 01   |          6 |
| 02   |          6 |
| 03   |          6 |
+------+------------+
3 rows in set (0.00 sec)

20.查询出只选修两门课程的学生学号和姓名

mysql> select student.sid, student.sname from student
    -> where student.sid in
    -> (select sc.sid from sc
    -> group by sc.sid
    -> having count(sc.cid)=2
    -> );

+------+--------+
| sid  | sname  |
+------+--------+
| 05   | 周梅   |
| 06   | 吴兰   |
| 07   | 郑竹   |
+------+--------+
3 rows in set (0.03 sec)