mysql - 经典50题

经典案例50题，确实经典，在我大学的时候，老师就把这些题目，当成课下作业给我们做了；

还有另一套题目，是算部门收入的，也非常经典，如果这些题不够做，可以再找找（建表语句包含employee的）；

这些题目包含多个答案，有一些实习生做的，添加了部分语法分析，可能仍有改进空间。

50题大纲

1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数 
2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数
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
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
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
6、查询"李"姓老师的数量 
7、查询学过"张三"老师授课的同学的信息 
8、查询没学过"张三"老师授课的同学的信息 
9、查询学过编号为"01"并且也学过编号为"02"的课程的同学的信息
10、查询学过编号为"01"但是没有学过编号为"02"的课程的同学的信息
11、查询没有学全所有课程的同学的信息
12、查询至少有一门课与学号为"01"的同学所学相同的同学的信息
13、查询和"01"号的同学学习的课程完全相同的其他同学的信息
14、查询没学过"张三"老师讲授的任一门课程的学生姓名
15、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
16、检索"01"课程分数小于60，按分数降序排列的学生信息
17、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
18、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
19、按各科成绩进行排序，并显示排名(实现不完全)
20、查询学生的总成绩并进行排名
21、查询不同老师所教不同课程平均分从高到低显示
22、查询所有课程的成绩第2名到第3名的学生信息及该课程成绩
23、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],[0-60]及所占百分比
24、查询学生平均成绩及其名次
25、查询各科成绩前三名的记录
26、查询每门课程被选修的学生数
27、查询出只有两门课程的全部学生的学号和姓名
28、查询男生、女生人数
29、查询名字中含有"风"字的学生信息
30、查询同名同性学生名单，并统计同名人数
31、查询1990年出生的学生名单
32、查询每门课程的平均成绩，结果按平均成绩降序排列，平均成绩相同时，按课程编号升序排列
33、查询平均成绩大于等于85的所有学生的学号、姓名和平均成绩
34、查询课程名称为"数学"，且分数低于60的学生姓名和分数
35、查询所有学生的课程及分数情况；
36、查询任何一门课程成绩在70分以上的姓名、课程名称和分数
37、查询不及格的课程
38、查询课程编号为01且课程成绩在80分以上的学生的学号和姓名
39、求每门课程的学生人数
40、查询选修"张三"老师所授课程的学生中，成绩最高的学生信息及其成绩
41、查询不同课程成绩相同的学生的学生编号、课程编号、学生成绩
42、查询每门功成绩最好的前两名
43、统计每门课程的学生选修人数（超过5人的课程才统计）；要求输出课程号和选修人数，查询结果按人数降序排列。
44、检索至少选修两门课程的学生学号 
45、查询选修了全部课程的学生信息
46、查询各学生的年龄
47、查询本周过生日的学生
48、查询下周过生日的学生
49、查询本月过生日的学生
50、查询下月过生日的学生

建表语句

/*
 navicat premium data transfer

 source server         : localhost
 source server type    : mysql
 source server version : 50744
 source host           : localhost:3306
 source schema         : sqlpractice

 target server type    : mysql
 target server version : 50744
 file encoding         : 65001

 date: 17/06/2024 17:24:13
*/

set names utf8mb4;
set foreign_key_checks = 0;

-- ----------------------------
-- table structure for course
-- ----------------------------
drop table if exists `course`;
create table `course`  (
  `c_id` varchar(20)  not null,
  `c_name` varchar(20)  not null,
  `t_id` varchar(20)  not null,
  primary key (`c_id`) using btree,
  index `t_id`(`t_id`) using btree,
  constraint `course_ibfk_1` foreign key (`t_id`) references `teacher` (`t_id`) on delete restrict on update restrict
) engine = innodb character set = utf8 collate = utf8_general_ci row_format = dynamic;

-- ----------------------------
-- records of course
-- ----------------------------
insert into `course` values ('01', '语文', '02');
insert into `course` values ('02', '数学', '01');
insert into `course` values ('03', '英语', '03');

-- ----------------------------
-- table structure for score
-- ----------------------------
drop table if exists `score`;
create table `score`  (
  `s_id` varchar(20)  not null,
  `c_id` varchar(20)  not null,
  `s_score` int(3) null default null,
  primary key (`s_id`, `c_id`) using btree,
  index `c_id`(`c_id`) using btree,
  constraint `score_ibfk_1` foreign key (`s_id`) references `student` (`s_id`) on delete restrict on update restrict,
  constraint `score_ibfk_2` foreign key (`c_id`) references `course` (`c_id`) on delete restrict on update restrict
) engine = innodb character set = utf8 collate = utf8_general_ci row_format = dynamic;

-- ----------------------------
-- records of score
-- ----------------------------
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);

-- ----------------------------
-- table structure for student
-- ----------------------------
drop table if exists `student`;
create table `student`  (
  `s_id` varchar(20)  not null,
  `s_name` varchar(20)  not null,
  `s_birth` varchar(20)  not null,
  `s_sex` varchar(10)  not null,
  primary key (`s_id`) using btree
) engine = innodb character set = utf8 collate = utf8_general_ci row_format = dynamic;

-- ----------------------------
-- records of student
-- ----------------------------
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');

-- ----------------------------
-- table structure for teacher
-- ----------------------------
drop table if exists `teacher`;
create table `teacher`  (
  `t_id` varchar(20)  not null,
  `t_name` varchar(20)  not null default '',
  primary key (`t_id`) using btree
) engine = innodb character set = utf8 collate = utf8_general_ci row_format = dynamic;

-- ----------------------------
-- records of teacher
-- ----------------------------
insert into `teacher` values ('01', '张三');
insert into `teacher` values ('02', '李四');
insert into `teacher` values ('03', '王五');

set foreign_key_checks = 1;

1. 查询“01”课程比“02”课程成绩高的学生的信息及课程分数。

-- 经典语法 join

-- 已知问题：
-- 如果去除 where 条件，总共 48 条记录，远大于任何一张基础表，
-- 生产环境这么查，大概率查不出数据
SELECT
	stu.*,
	s1.s_score 
FROM
	student AS stu
	INNER JOIN score AS s1 ON stu.s_id = s1.s_id
	INNER JOIN score AS s2 ON s1.s_id = s2.s_id 	
WHERE
	s1.s_score > s2.s_score 
	AND s1.c_id = '01'
	AND s2.c_id = '02';


-- 使用子查询嵌套，提前过滤出数据，之后再进行 join
SELECT
	stu.*,
	s1.s_score 
FROM
	student AS stu
	INNER JOIN (SELECT * FROM score where score.c_id = '01') AS s1 ON stu.s_id = s1.s_id
	INNER JOIN (SELECT * FROM score where score.c_id = '02') AS s2 ON s1.s_id = s2.s_id 	
WHERE
	s1.s_score > s2.s_score;

-- 或者这么写
SELECT
	stu.*,
	s1.c_id,
	s1.s_score,
	s2.c_id, 
	s2.s_score
FROM
	student AS stu
	INNER JOIN score AS s1 ON stu.s_id = s1.s_id AND s1.c_id = '01'
	INNER JOIN score AS s2 ON stu.s_id = s2.s_id AND s2.c_id = '02'	
WHERE
	s1.s_score > s2.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。（同上）

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。

思路： 求平均成绩可用聚合函数，考虑表关联直接分组。

-- 经典语法：group by ... having ...
SELECT
	s.s_id,
	s.s_name,
	AVG(sc.s_score) avg_score
FROM
	Student s
JOIN Score sc ON s.s_id = sc.s_id
GROUP BY
	s.s_id,
	s.s_name
HAVING
	avg_score >= 60


-- 使用子查询
SELECT
	stu.s_id,
	stu.s_name,
	avg.avg_score
FROM
	student AS stu
INNER JOIN (
	SELECT
		s_id,
		avg(s_score) AS avg_score
	FROM
		score
	GROUP BY
		s_id
) AS avg ON stu.s_id = avg.s_id
WHERE 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）。

SELECT
	s.s_id,
	s.s_name,
	AVG(sc.s_score) avg_score
FROM
	Student s
LEFT JOIN Score sc ON s.s_id = sc.s_id
GROUP BY
	s.s_id,
	s.s_name
HAVING
	avg_score < 60
OR avg_score IS NULL

-- 如果业务允许，也可以用 ifnull 替换空值
SELECT
	s.s_id,
	s.s_name,
	AVG(IFNULL(sc.s_score, 0)) avg_score
FROM
	Student s
LEFT JOIN Score sc ON s.s_id = sc.s_id
GROUP BY
	s.s_id,
	s.s_name
HAVING
	avg_score < 60

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。（同案例3）

6. 查询“李”姓老师的数量。

-- 基础语法
SELECT COUNT(*)
FROM teacher
WHERE t_name LIKE '李%' 

7. 查询学过张三老师授课的同学的信息。

-- 同案例1
SELECT
	stu.* 
FROM
	student AS stu
	INNER JOIN score ON stu.s_id = score.s_id
	INNER JOIN course ON score.c_id = course.c_id
	INNER JOIN teacher ON course.t_id = teacher.t_id 
WHERE
	t_name = '张三';

8. 找出没有学过张三老师课程的学生

思路： 参考题目7查询出张三老师授课的学生信息将他们排除。

-- 方式 1：经典 not in 写法
-- 已知问题：not in 的子查询本部分，如果结果集太大，会导致内存相关的报错
SELECT
	*
FROM
	student 
where student.s_id not in (
	SELECT 
		sc.s_id,
		c.c_name,
		t.t_name
	FROM
		score sc
	JOIN course c ON sc.c_id = c.c_id
	JOIN teacher t ON c.t_id = t.t_id
	AND t_name = '张三'
)

-- 方式 2：用 join 代替 not in
-- 写法通常优于上一句，但是 is null 条件，也用不到索引
select a.* from student a
left join (
	SELECT 
		sc.s_id,
		c.c_name,
		t.t_name
	FROM
		score sc
	JOIN course c ON sc.c_id = c.c_id
	JOIN teacher t ON c.t_id = t.t_id
	AND t_name = '张三'
) b on a.s_id = b.s_id
where b.s_id is null

9. 查询学过编号为 01，并且学过编号为 02 课程的学生信息

-- 使用 in 语法
SELECT s_id, s_name
FROM Student
WHERE s_id IN(
  SELECT s1.s_id
  FROM Score s1
  JOIN Score s2
  ON s1.s_id = s2.s_id
  WHERE s1.c_id = '01' AND s2.c_id = '02'
)

-- 使用 join 语法
SELECT
	stu.*,
	score1.s_id,
	score2.s_id
FROM
	student AS stu
	LEFT JOIN score AS score1 ON stu.s_id = score1.s_id and score1.c_id = '01' 
	LEFT JOIN score AS score2 ON stu.s_id = score2.s_id and score2.c_id = '02'
where score1.s_id is not null and score2.s_id is not null;


-- 使用 join 语法（不太好的写法）	
SELECT
	stu.* 
FROM
	student AS stu
	LEFT JOIN score AS score1 ON stu.s_id = score1.s_id
	LEFT JOIN score AS score2 ON score1.s_id = score2.s_id 
WHERE
	score1.c_id = '01' 
	AND score2.c_id = '02';

10. 查询学过 01 课程，但是没有学过 02 课程的学生信息

-- 使用 in 语法
SELECT
	s.*
FROM
	student s
JOIN score sc ON s.s_id = sc.s_id
WHERE
	sc.c_id = '01'
AND s.s_id NOT IN (
	SELECT
		s_id
	FROM
		score
	WHERE
		c_id = '02'
)

-- 使用 join 语法
SELECT
	stu.*,
	score1.s_id,
	score2.s_id
FROM
	student AS stu
	LEFT JOIN score AS score1 ON stu.s_id = score1.s_id and score1.c_id = '01' 
	LEFT JOIN score AS score2 ON stu.s_id = score2.s_id and score2.c_id = '02'
where score1.s_id is not null and score2.s_id is null;

待续……