mysql之魔鬼训练营

普通 + 中等 难度练习题

测试数据：

--建表
--学生表
CREATE TABLE `Student`(
	`s_id` VARCHAR(20),
	`s_name` VARCHAR(20) NOT NULL DEFAULT '',
	`s_birth` VARCHAR(20) NOT NULL DEFAULT '',
	`s_sex` VARCHAR(10) NOT NULL DEFAULT '',
	PRIMARY KEY(`s_id`)
);
--课程表
CREATE TABLE `Course`(
	`c_id`  VARCHAR(20),
	`c_name` VARCHAR(20) NOT NULL DEFAULT '',
	`t_id` VARCHAR(20) NOT NULL,
	PRIMARY KEY(`c_id`)
);
--教师表
CREATE TABLE `Teacher`(
	`t_id` VARCHAR(20),
	`t_name` VARCHAR(20) NOT NULL DEFAULT '',
	PRIMARY KEY(`t_id`)
);
--成绩表
CREATE TABLE `Score`(
	`s_id` VARCHAR(20),
	`c_id`  VARCHAR(20),
	`s_score` INT(3),
	PRIMARY KEY(`s_id`,`c_id`)
);
--插入学生表测试数据
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' , '1990-05-20' , '男');
insert into Student values('04' , '李云' , '1990-08-06' , '男');
insert into Student values('05' , '周梅' , '1991-12-01' , '女');
insert into Student values('06' , '吴兰' , '1992-03-01' , '女');
insert into Student values('07' , '郑竹' , '1989-07-01' , '女');
insert into Student values('08' , '王菊' , '1990-01-20' , '女');
--课程表测试数据
insert into Course values('01' , '语文' , '02');
insert into Course values('02' , '数学' , '01');
insert into Course values('03' , '英语' , '03');

--教师表测试数据
insert into Teacher values('01' , '张三');
insert into Teacher values('02' , '李四');
insert into Teacher values('03' , '王五');

--成绩表测试数据
insert into Score values('01' , '01' , 80);
insert into Score values('01' , '02' , 90);
insert into Score values('01' , '03' , 99);
insert into Score values('02' , '01' , 70);
insert into Score values('02' , '02' , 60);
insert into Score values('02' , '03' , 80);
insert into Score values('03' , '01' , 80);
insert into Score values('03' , '02' , 80);
insert into Score values('03' , '03' , 80);
insert into Score values('04' , '01' , 50);
insert into Score values('04' , '02' , 30);
insert into Score values('04' , '03' , 20);
insert into Score values('05' , '01' , 76);
insert into Score values('05' , '02' , 87);
insert into Score values('06' , '01' , 31);
insert into Score values('06' , '03' , 34);
insert into Score values('07' , '02' , 89);
insert into Score values('07' , '03' , 98);

视图展示：

# 学生表
mysql> select * from student;
+------+--------+------------+-------+
| s_id | s_name | s_birth    | s_sex |
+------+--------+------------+-------+
| 01   | 赵雷   | 1990-01-01 | 男    |
| 02   | 钱电   | 1990-12-21 | 男    |
| 03   | 孙风   | 1990-05-20 | 男    |
| 04   | 李云   | 1990-08-06 | 男    |
| 05   | 周梅   | 1991-12-01 | 女    |
| 06   | 吴兰   | 1992-03-01 | 女    |
| 07   | 郑竹   | 1989-07-01 | 女    |
| 08   | 王菊   | 1990-01-20 | 女    |
+------+--------+------------+-------+
8 rows in set (0.00 sec)

# 老师表
mysql> select * from teacher;
+------+--------+
| t_id | t_name |
+------+--------+
| 01   | 张三   |
| 02   | 李四   |
| 03   | 王五   |
+------+--------+
3 rows in set (0.00 sec)

# 课程表
mysql> select * from course;
+------+--------+------+
| c_id | c_name | t_id |
+------+--------+------+
| 01   | 语文   | 02   |
| 02   | 数学   | 01   |
| 03   | 英语   | 03   |
+------+--------+------+
3 rows in set (0.00 sec)

# 成绩表
mysql> select * from score;
+------+------+---------+
| s_id | c_id | s_score |
+------+------+---------+
| 01   | 01   |      80 |
| 01   | 02   |      90 |
| 01   | 03   |      99 |
| 02   | 01   |      70 |
| 02   | 02   |      60 |
| 02   | 03   |      80 |
| 03   | 01   |      80 |
| 03   | 02   |      80 |
| 03   | 03   |      80 |
| 04   | 01   |      50 |
| 04   | 02   |      30 |
| 04   | 03   |      20 |
| 05   | 01   |      76 |
| 05   | 02   |      87 |
| 06   | 01   |      31 |
| 06   | 03   |      34 |
| 07   | 02   |      89 |
| 07   | 03   |      98 |
+------+------+---------+
18 rows in set (0.00 sec)

练习题：

# 1. 查询“01”课程比“02”课程成绩高的学生的信息及课程分数
# a代表学生表（student）  b代表成绩表（score）
select a.*,b.s_score as 01_score ,c.s_score as 02_score from
student a join score b on b.s_id = a.s_id and b.c_id='01' left join score c on a.s_id=c.s_id and c.c_id="02" or c.c_id = NULL where b.s_score>c.s_score;

# 答案：
select a.* ,b.s_score as 01_score,c.s_score as 02_score from
	student a join score b on a.s_id=b.s_id and b.c_id='01'
		left join score c on a.s_id=c.s_id and c.c_id='02' or c.c_id = NULL where b.s_score>c.s_score;

# 2. 查询"01"课程比"02"课程成绩低的学生的信息及课程分数
# a代表学生表（student）  b代表成绩表（score）  c代表成绩表（score）
select a.*,b.s_score as score01,c.s_score as score02 from
	student a join score b on a.s_id = b.s_id and b.c_id = "01"
	join score c on a.s_id = c.s_id and c.c_id="02" where c.s_score > b.s_score;

# 答案
select a.* ,b.s_score as 01_score,c.s_score as 02_score from
	student a left join score b on a.s_id=b.s_id and b.c_id='01' or b.c_id=NULL
	 join score c on a.s_id=c.s_id and c.c_id='02' where b.s_score<c.s_score;

# 3. 查询平均成绩大于等于60分的同学的学生编号和学生姓名和平均成绩
# a代表学生表（student）  b代表成绩表（score）  c代表成绩表（score）
select a.s_id,a.s_name,AVG(b.s_score) as average  from student a join score b on a.s_id = b.s_id group by b.s_id  HAVING average>60;

# 答案
select b.s_id,b.s_name,ROUND(AVG(a.s_score),2) as avg_score from
	student b
	join score a on b.s_id = a.s_id
	GROUP BY b.s_id,b.s_name HAVING avg_score >=60;

# 4. 查询平均成绩小于60分的同学的学生编号和学生姓名和平均成绩-- (包括有成绩的和无成绩的)
# a代表学生表（student） b代表成绩表 (score)
select a.s_id,a.s_name,AVG(b.s_score) as average  from student a  left join score b on a.s_id = b.s_id  group by b.s_id  HAVING average < 60 

# 答案：
select b.s_id,b.s_name,ROUND(AVG(a.s_score),2) as avg_score from
	student b
	left join score a on b.s_id = a.s_id
	GROUP BY b.s_id,b.s_name HAVING avg_score <60
	union
select a.s_id,a.s_name,0 as avg_score from
	student a
	where a.s_id not in (
				select distinct s_id from score);

# 5.查询所有同学的学生编号、学生姓名、选课总数、所有课程的总成绩
# a代表学生表（student） b代表课程表 (score)
select a.s_id,a.s_name,count(b.c_id) as sum_course,SUM(b.s_score) as sum_score from student a left join score b on a.s_id = b.s_id group by a.s_id,a.s_name;

# 答案
select a.s_id,a.s_name,count(b.c_id) as sum_course,sum(b.s_score) as sum_score from
	student a
	left join score b on a.s_id=b.s_id
	GROUP BY a.s_id,a.s_name;

# 6.查询"李"姓老师的数量
select count(*) from teacher  where t_name like "李%"; 

# 答案：
select count(t_id) from teacher where t_name like '李%';