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<b.score and a.cid = b.cid
-> 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<a.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
-> );