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数据库系统原理与应用教程（076）—— MySQL 练习题：操作题 160-167（二十）：综合练习
2022-08-03 11:55:00 【睿思达DBA_WGX】
数据库系统原理与应用教程（076）—— MySQL 练习题：操作题 160-167（二十）：综合练习
160、聚合函数的使用（1）
该题目使用的表和数据如下：
/* drop table if exists resume_info; CREATE TABLE resume_info ( id int(4) NOT NULL, job varchar(64) NOT NULL, date date NOT NULL, num int(11) NOT NULL, PRIMARY KEY (id)); INSERT INTO resume_info VALUES (1,'C++','2025-01-02',53), (2,'Python','2025-01-02',23), (3,'Java','2025-01-02',12), (4,'Java','2025-02-03',24), (5,'C++','2025-02-03',23), (6,'Python','2025-02-03',34), (7,'Python','2025-03-04',54), (8,'C++','2025-03-04',65), (9,'Java','2025-03-04',92), (10,'Java','2026-01-04',230); */
简历信息表：resume_info，表中数据如下：
mysql> select * from resume_info;
+----+--------+------------+-----+
| id | job | date | num |
+----+--------+------------+-----+
| 1 | C++ | 2025-01-02 | 53 |
| 2 | Python | 2025-01-02 | 23 |
| 3 | Java | 2025-01-02 | 12 |
| 4 | Java | 2025-02-03 | 24 |
| 5 | C++ | 2025-02-03 | 23 |
| 6 | Python | 2025-02-03 | 34 |
| 7 | Python | 2025-03-04 | 54 |
| 8 | C++ | 2025-03-04 | 65 |
| 9 | Java | 2025-03-04 | 92 |
| 10 | Java | 2026-01-04 | 230 |
+----+--------+------------+-----+
10 rows in set (0.00 sec)
【问题】请编写 SQL 语句，查询在 2025 年内投递简历的岗位和数量，并且按数量降序排序，查询结果如下：
| job | cnt |
|---|---|
| C++ | 141 |
| Java | 128 |
| Python | 111 |
解答：
/* select job, sum(num) cnt from resume_info where date between '2025-1-1' and '2025-12-31' group by job order by cnt desc; */
mysql> select job, sum(num) cnt
-> from resume_info where date between '2025-1-1' and '2025-12-31'
-> group by job
-> order by cnt desc;
+--------+------+
| job | cnt |
+--------+------+
| C++ | 141 |
| Java | 128 |
| Python | 111 |
+--------+------+
3 rows in set (0.00 sec)
161、聚合函数的使用（2）
该题目使用的表和数据如下：
/* drop table if exists resume_info; CREATE TABLE resume_info ( id int(4) NOT NULL, job varchar(64) NOT NULL, date date NOT NULL, num int(11) NOT NULL, PRIMARY KEY (id)); INSERT INTO resume_info VALUES (1,'C++','2025-01-02',53), (2,'Python','2025-01-02',23), (3,'Java','2025-01-02',12), (4,'C++','2025-01-03',54), (5,'Python','2025-01-03',43), (6,'Java','2025-01-03',41), (7,'Java','2025-02-03',24), (8,'C++','2025-02-03',23), (9,'Python','2025-02-03',34), (10,'Java','2025-02-04',42), (11,'C++','2025-02-04',45), (12,'Python','2025-02-04',59), (13,'Python','2025-03-04',54), (14,'C++','2025-03-04',65), (15,'Java','2025-03-04',92), (16,'Python','2025-03-05',34), (17,'C++','2025-03-05',34), (18,'Java','2025-03-05',34), (19,'Python','2026-01-04',230), (20,'C++','2026-02-06',231); */
简历信息表：resume_info，表中数据如下：
mysql> select * from resume_info;
+----+--------+------------+-----+
| id | job | date | num |
+----+--------+------------+-----+
| 1 | C++ | 2025-01-02 | 53 |
| 2 | Python | 2025-01-02 | 23 |
| 3 | Java | 2025-01-02 | 12 |
| 4 | C++ | 2025-01-03 | 54 |
| 5 | Python | 2025-01-03 | 43 |
| 6 | Java | 2025-01-03 | 41 |
| 7 | Java | 2025-02-03 | 24 |
| 8 | C++ | 2025-02-03 | 23 |
| 9 | Python | 2025-02-03 | 34 |
| 10 | Java | 2025-02-04 | 42 |
| 11 | C++ | 2025-02-04 | 45 |
| 12 | Python | 2025-02-04 | 59 |
| 13 | Python | 2025-03-04 | 54 |
| 14 | C++ | 2025-03-04 | 65 |
| 15 | Java | 2025-03-04 | 92 |
| 16 | Python | 2025-03-05 | 34 |
| 17 | C++ | 2025-03-05 | 34 |
| 18 | Java | 2025-03-05 | 34 |
| 19 | Python | 2026-01-04 | 230 |
| 20 | C++ | 2026-02-06 | 231 |
+----+--------+------------+-----+
20 rows in set (0.00 sec)
【问题】请编写 SQL 语句，查询在 2025 年内投递简历的每个岗位，每一个月内收到简历的数量，先按月份降序排序，再按简历数目降序排序，查询结果如下:
| job | mon | cnt |
|---|---|---|
| Java | 2025-03 | 126 |
| C++ | 2025-03 | 99 |
| Python | 2025-03 | 88 |
| Python | 2025-02 | 93 |
| C++ | 2025-02 | 68 |
| Java | 2025-02 | 66 |
| C++ | 2025-01 | 107 |
| Python | 2025-01 | 66 |
| Java | 2025-01 | 53 |
/* select job, concat(left(date, 4), '-', substr(date, 6, 2)) mon, sum(num) cnt from resume_info where date between '2025-1-1' and '2025-12-31' group by job, mon order by mon desc, cnt desc; */
mysql> select job,
-> concat(left(date, 4), '-', substr(date, 6, 2)) mon,
-> sum(num) cnt
-> from resume_info
-> where date between '2025-1-1' and '2025-12-31'
-> group by job, mon
-> order by mon desc, cnt desc;
+--------+---------+------+
| job | mon | cnt |
+--------+---------+------+
| Java | 2025-03 | 126 |
| C++ | 2025-03 | 99 |
| Python | 2025-03 | 88 |
| Python | 2025-02 | 93 |
| C++ | 2025-02 | 68 |
| Java | 2025-02 | 66 |
| C++ | 2025-01 | 107 |
| Python | 2025-01 | 66 |
| Java | 2025-01 | 53 |
+--------+---------+------+
9 rows in set (0.00 sec)
162、聚合函数的使用（3）
该题目使用的表和数据如下：
/* drop table if exists resume_info; CREATE TABLE resume_info ( id int(4) NOT NULL, job varchar(64) NOT NULL, date date NOT NULL, num int(11) NOT NULL, PRIMARY KEY (id)); INSERT INTO resume_info VALUES (1,'C++','2025-01-02',53), (2,'Python','2025-01-02',23), (3,'Java','2025-01-02',12), (4,'C++','2025-01-03',54), (5,'Python','2025-01-03',43), (6,'Java','2025-01-03',41), (7,'Java','2025-02-03',24), (8,'C++','2025-02-03',23), (9,'Python','2025-02-03',34), (10,'Java','2025-02-04',42), (11,'C++','2025-02-04',45), (12,'Python','2025-02-04',59), (13,'C++','2026-01-04',230), (14,'Java','2026-01-04',764), (15,'Python','2026-01-04',644), (16,'C++','2026-01-06',240), (17,'Java','2026-01-06',714), (18,'Python','2026-01-06',624), (19,'C++','2026-02-14',260), (20,'Java','2026-02-14',721), (21,'Python','2026-02-14',321), (22,'C++','2026-02-24',134), (23,'Java','2026-02-24',928), (24,'Python','2026-02-24',525), (25,'C++','2027-02-06',231); */
简历信息表：resume_info，表中数据如下：
mysql> select * from resume_info;
+----+--------+------------+-----+
| id | job | date | num |
+----+--------+------------+-----+
| 1 | C++ | 2025-01-02 | 53 |
| 2 | Python | 2025-01-02 | 23 |
| 3 | Java | 2025-01-02 | 12 |
| 4 | C++ | 2025-01-03 | 54 |
| 5 | Python | 2025-01-03 | 43 |
| 6 | Java | 2025-01-03 | 41 |
| 7 | Java | 2025-02-03 | 24 |
| 8 | C++ | 2025-02-03 | 23 |
| 9 | Python | 2025-02-03 | 34 |
| 10 | Java | 2025-02-04 | 42 |
| 11 | C++ | 2025-02-04 | 45 |
| 12 | Python | 2025-02-04 | 59 |
| 13 | C++ | 2026-01-04 | 230 |
| 14 | Java | 2026-01-04 | 764 |
| 15 | Python | 2026-01-04 | 644 |
| 16 | C++ | 2026-01-06 | 240 |
| 17 | Java | 2026-01-06 | 714 |
| 18 | Python | 2026-01-06 | 624 |
| 19 | C++ | 2026-02-14 | 260 |
| 20 | Java | 2026-02-14 | 721 |
| 21 | Python | 2026-02-14 | 321 |
| 22 | C++ | 2026-02-24 | 134 |
| 23 | Java | 2026-02-24 | 928 |
| 24 | Python | 2026-02-24 | 525 |
| 25 | C++ | 2027-02-06 | 231 |
+----+--------+------------+-----+
25 rows in set (0.00 sec)
【问题】请编写 SQL 语句，查询在 2025 年投递简历的每个岗位，每一个月内收到简历的数目，和对应的 2026 年的同一个月同岗位，收到简历的数目，最后的结果先按 first_year_mon 月份降序，再按 job 降序排序显示，查询结果如下：
| job | first_year_mon | first_year_cnt | second_year_mon | second_year_cnt |
|---|---|---|---|---|
| Python | 2025-02 | 93 | 2026-02 | 846 |
| Java | 2025-02 | 66 | 2026-02 | 1649 |
| C++ | 2025-02 | 68 | 2026-02 | 394 |
| Python | 2025-01 | 66 | 2026-01 | 1268 |
| Java | 2025-01 | 53 | 2026-01 | 1478 |
| C++ | 2025-01 | 107 | 2026-01 | 470 |
解答：
/* select a.job, a.mon first_year_mon, a.cnt first_year_cnt, b.mon second_year_mon, b.cnt second_year_cnt from (select job, concat(left(date, 4), '-', substr(date, 6, 2)) mon, sum(num) cnt from resume_info where date between '2025-1-1' and '2025-12-31' group by job, mon order by mon desc, cnt desc) a join (select job, concat(left(date, 4), '-', substr(date, 6, 2)) mon, sum(num) cnt from resume_info where date between '2026-1-1' and '2026-12-31' group by job, mon order by mon desc, cnt desc) b on a.job = b.job and right(a.mon, 2) = right(b.mon, 2) order by first_year_mon desc, job desc; */
mysql> select a.job, a.mon first_year_mon, a.cnt first_year_cnt,
-> b.mon second_year_mon, b.cnt second_year_cnt
-> from
-> (select job,
-> concat(left(date, 4), '-', substr(date, 6, 2)) mon,
-> sum(num) cnt
-> from resume_info
-> where date between '2025-1-1' and '2025-12-31'
-> group by job, mon
-> order by mon desc, cnt desc) a join
-> (select job,
-> concat(left(date, 4), '-', substr(date, 6, 2)) mon,
-> sum(num) cnt
-> from resume_info
-> where date between '2026-1-1' and '2026-12-31'
-> group by job, mon
-> order by mon desc, cnt desc) b
-> on a.job = b.job and right(a.mon, 2) = right(b.mon, 2)
-> order by first_year_mon desc, job desc;
+--------+----------------+----------------+-----------------+-----------------+
| job | first_year_mon | first_year_cnt | second_year_mon | second_year_cnt |
+--------+----------------+----------------+-----------------+-----------------+
| Python | 2025-02 | 93 | 2026-02 | 846 |
| Java | 2025-02 | 66 | 2026-02 | 1649 |
| C++ | 2025-02 | 68 | 2026-02 | 394 |
| Python | 2025-01 | 66 | 2026-01 | 1268 |
| Java | 2025-01 | 53 | 2026-01 | 1478 |
| C++ | 2025-01 | 107 | 2026-01 | 470 |
+--------+----------------+----------------+-----------------+-----------------+
6 rows in set (0.01 sec)
163、聚合函数的使用（4）
该题目使用的表和数据如下：
/* drop table if exists user; drop table if exists grade_info; CREATE TABLE user ( id int(4) NOT NULL, name varchar(32) NOT NULL ); CREATE TABLE grade_info ( user_id int(4) NOT NULL, grade_num int(4) NOT NULL, type varchar(32) NOT NULL ); INSERT INTO user VALUES (1,'tm'), (2,'wwy'), (3,'zk'), (4,'qq'), (5,'lm'); INSERT INTO grade_info VALUES (1,3,'add'), (2,3,'add'), (1,1,'add'), (3,3,'add'), (4,3,'add'), (5,3,'add'); */
用户表：user，表中数据如下：
mysql> select * from user;
+----+------+
| id | name |
+----+------+
| 1 | tm |
| 2 | wwy |
| 3 | zk |
| 4 | qq |
| 5 | lm |
+----+------+
5 rows in set (0.00 sec)
积分表：grade_info，表中数据如下：
mysql> select * from grade_info;
+---------+-----------+------+
| user_id | grade_num | type |
+---------+-----------+------+
| 1 | 3 | add |
| 2 | 3 | add |
| 1 | 1 | add |
| 3 | 3 | add |
| 4 | 3 | add |
| 5 | 3 | add |
+---------+-----------+------+
6 rows in set (0.00 sec)
【问题】请编写一个 SQL 语句，查找积分增加最高的用户的名字，以及他的总积分是多少，查询结果如下：
| name | grade_num |
|---|---|
| tm | 4 |
解答：
/* select u.name, a.grade_num from (select user_id, sum(grade_num) grade_num from grade_info group by user_id order by grade_num desc limit 1) a join user u on a.user_id = u.id; */
mysql> select u.name, a.grade_num
-> from (select user_id, sum(grade_num) grade_num
-> from grade_info
-> group by user_id
-> order by grade_num desc limit 1) a join user u
-> on a.user_id = u.id;
+------+-----------+
| name | grade_num |
+------+-----------+
| tm | 4 |
+------+-----------+
1 row in set (0.00 sec)
164、聚合函数的使用（5）
该题目使用的表和数据如下：
/* drop table if exists user; drop table if exists grade_info; CREATE TABLE user ( id int(4) NOT NULL, name varchar(32) NOT NULL ); CREATE TABLE grade_info ( user_id int(4) NOT NULL, grade_num int(4) NOT NULL, type varchar(32) NOT NULL ); INSERT INTO user VALUES (1,'tm'), (2,'wwy'), (3,'zk'), (4,'qq'), (5,'lm'); INSERT INTO grade_info VALUES (1,3,'add'), (2,3,'add'), (1,1,'add'), (3,3,'add'), (4,3,'add'), (5,3,'add'), (3,1,'add'); */
用户表：user，表中数据如下：
mysql> select * from user;
+----+------+
| id | name |
+----+------+
| 1 | tm |
| 2 | wwy |
| 3 | zk |
| 4 | qq |
| 5 | lm |
+----+------+
5 rows in set (0.00 sec)
积分表：grade_info，表中数据如下：
mysql> select * from grade_info;
+---------+-----------+------+
| user_id | grade_num | type |
+---------+-----------+------+
| 1 | 3 | add |
| 2 | 3 | add |
| 1 | 1 | add |
| 3 | 3 | add |
| 4 | 3 | add |
| 5 | 3 | add |
| 3 | 1 | add |
+---------+-----------+------+
7 rows in set (0.00 sec)
【问题】请编写 SQL 语句，查找积分增加最高的用户的 id，名字以及总积分，查询结果按照 id 升序排序，查询结果如下：
| id | name | grade_num |
|---|---|---|
| 1 | tm | 4 |
| 3 | zk | 4 |
解答：
/* select u.id, u.name, sum(if(gi.type = 'add', grade_num, - grade_num)) grade_num from user u join grade_info gi on u.id = gi.user_id group by u.id, u.name having grade_num in (select max(grade_num) from (select sum(if(type = 'add', grade_num, - grade_num)) grade_num from grade_info group by user_id) a) order by u.id; */
mysql> select u.id, u.name, sum(grade_num) grade_num
-> from user u join grade_info gi on u.id = gi.user_id
-> group by u.id, u.name
-> having grade_num in
-> (select max(grade_num) from
-> (select sum(grade_num) grade_num from grade_info group by user_id) a)
-> order by u.id;
+----+------+-----------+
| id | name | grade_num |
+----+------+-----------+
| 1 | tm | 4 |
| 3 | zk | 4 |
+----+------+-----------+
2 rows in set (0.00 sec)
165、聚合函数的使用（6）
该题目使用的表和数据如下：
/* drop table if exists user; drop table if exists grade_info; CREATE TABLE user ( id int(4) NOT NULL, name varchar(32) NOT NULL ); CREATE TABLE grade_info ( user_id int(4) NOT NULL, grade_num int(4) NOT NULL, type varchar(32) NOT NULL ); INSERT INTO user VALUES (1,'tm'), (2,'wwy'), (3,'zk'), (4,'qq'), (5,'lm'); INSERT INTO grade_info VALUES (1,3,'add'), (2,3,'add'), (1,1,'reduce'), (3,3,'add'), (4,3,'add'), (5,3,'add'), (3,1,'reduce'); */
用户表：user，表中数据如下：
mysql> select * from user;
+----+------+
| id | name |
+----+------+
| 1 | tm |
| 2 | wwy |
| 3 | zk |
| 4 | qq |
| 5 | lm |
+----+------+
5 rows in set (0.00 sec)
积分表：grade_info，表中数据如下：
mysql> select * from grade_info;
+---------+-----------+--------+
| user_id | grade_num | type |
+---------+-----------+--------+
| 1 | 3 | add |
| 2 | 3 | add |
| 1 | 1 | reduce |
| 3 | 3 | add |
| 4 | 3 | add |
| 5 | 3 | add |
| 3 | 1 | reduce |
+---------+-----------+--------+
7 rows in set (0.00 sec)
【问题】请编写一个 SQL 语句，查找积分增加最高的用户的名字，以及他的总积分是多少，查询结果如下：
| id | name | grade_num |
|---|---|---|
| 2 | wwy | 3 |
| 4 | 3 | |
| 5 | lm | 3 |
解答：
/* select u.id, u.name, sum(if(gi.type = 'add', grade_num, - grade_num)) grade_num from user u join grade_info gi on u.id = gi.user_id group by u.id, u.name having grade_num in (select max(grade_num) from (select sum(if(type = 'add', grade_num, - grade_num)) grade_num from grade_info group by user_id) a) order by u.id; */
mysql> select u.id, u.name, sum(if(gi.type = 'add', grade_num, - grade_num)) grade_num
-> from user u join grade_info gi on u.id = gi.user_id
-> group by u.id, u.name
-> having grade_num in
-> (select max(grade_num) from
-> (select sum(if(type = 'add', grade_num, - grade_num)) grade_num
-> from grade_info group by user_id) a)
-> order by u.id;
+----+------+-----------+
| id | name | grade_num |
+----+------+-----------+
| 2 | wwy | 3 |
| 4 | qq | 3 |
| 5 | lm | 3 |
+----+------+-----------+
3 rows in set (0.00 sec)
166、聚合函数的使用（7）
该题目使用的表和数据如下：
/* CREATE TABLE `goods` ( `id` int(11) NOT NULL, `name` varchar(10) DEFAULT NULL, `weight` int(11) NOT NULL, PRIMARY KEY (`id`) ); CREATE TABLE `trans` ( `id` int(11) NOT NULL, `goods_id` int(11) NOT NULL, `count` int(11) NOT NULL, PRIMARY KEY (`id`) ); insert into goods values(1,'A1',100); insert into goods values(2,'A2',20); insert into goods values(3,'B3',29); insert into goods values(4,'T1',60); insert into goods values(5,'G2',33); insert into goods values(6,'C0',55); insert into trans values(1,3,10); insert into trans values(2,1,44); insert into trans values(3,6,9); insert into trans values(4,1,2); insert into trans values(5,2,65); insert into trans values(6,5,23); insert into trans values(7,3,20); insert into trans values(8,2,16); insert into trans values(9,4,5); insert into trans values(10,1,3); */
商品表：goods，字段依次为：商品id，商品名，商品质量。表中数据如下：
mysql> select * from goods;
+----+------+--------+
| id | name | weight |
+----+------+--------+
| 1 | A1 | 100 |
| 2 | A2 | 20 |
| 3 | B3 | 29 |
| 4 | T1 | 60 |
| 5 | G2 | 33 |
| 6 | C0 | 55 |
+----+------+--------+
6 rows in set (0.00 sec)
交易表：trans，字段依次为：交易id，商品id，商品购买个数。表中数据如下：
mysql> select * from trans;
+----+----------+-------+
| id | goods_id | count |
+----+----------+-------+
| 1 | 3 | 10 |
| 2 | 1 | 44 |
| 3 | 6 | 9 |
| 4 | 1 | 2 |
| 5 | 2 | 65 |
| 6 | 5 | 23 |
| 7 | 3 | 20 |
| 8 | 2 | 16 |
| 9 | 4 | 5 |
| 10 | 1 | 3 |
+----+----------+-------+
10 rows in set (0.00 sec)
【问题】查询购买个数超过 20、质量小于 50 的商品，按照商品 id 升序排序，查询结果如下：
| id | name | weight | total |
|---|---|---|---|
| 2 | A2 | 20 | 81 |
| 3 | B3 | 29 | 30 |
| 5 | G2 | 33 | 23 |
解答：
/* select g.id, g.name, g.weight, sum(count) total from trans t join goods g on t.goods_id = g.id where g.weight < 50 group by g.id, g.name, g.weight having total > 20 order by g.id; */
mysql> select g.id, g.name, g.weight, sum(count) total
-> from trans t join goods g on t.goods_id = g.id
-> where g.weight < 50
-> group by g.id, g.name, g.weight
-> having total > 20
-> order by g.id;
+----+------+--------+-------+
| id | name | weight | total |
+----+------+--------+-------+
| 2 | A2 | 20 | 81 |
| 3 | B3 | 29 | 30 |
| 5 | G2 | 33 | 23 |
+----+------+--------+-------+
3 rows in set (0.00 sec)
167、连接查询
该题目使用的表和数据如下：
/* CREATE TABLE `follow` ( `user_id` int(4) NOT NULL, `follower_id` int(4) NOT NULL, PRIMARY KEY (`user_id`,`follower_id`)); CREATE TABLE `music_likes` ( `user_id` int(4) NOT NULL, `music_id` int(4) NOT NULL, PRIMARY KEY (`user_id`,`music_id`)); CREATE TABLE `music` ( `id` int(4) NOT NULL, `music_name` varchar(32) NOT NULL, PRIMARY KEY (`id`)); INSERT INTO follow VALUES(1,2); INSERT INTO follow VALUES(1,4); INSERT INTO follow VALUES(2,3); INSERT INTO music_likes VALUES(1,17); INSERT INTO music_likes VALUES(2,18); INSERT INTO music_likes VALUES(2,19); INSERT INTO music_likes VALUES(3,20); INSERT INTO music_likes VALUES(4,17); INSERT INTO music VALUES(17,'yueyawang'); INSERT INTO music VALUES(18,'kong'); INSERT INTO music VALUES(19,'MOM'); INSERT INTO music VALUES(20,'Sold Out'); */
关注表：follow，第一列是关注人的 id，第二列是被关注人的 id，这 2 列的 id 组成主键，表中数据如下：
mysql> select * from follow;
+---------+-------------+
| user_id | follower_id |
+---------+-------------+
| 1 | 2 |
| 1 | 4 |
| 2 | 3 |
+---------+-------------+
3 rows in set (0.00 sec)
个人的喜欢的音乐表：music_likes，第一列是用户 id，第二列是喜欢的音乐 id，这 2 列的 id 组成主键，表中数据如下：
mysql> select * from music_likes;
+---------+----------+
| user_id | music_id |
+---------+----------+
| 1 | 17 |
| 2 | 18 |
| 2 | 19 |
| 3 | 20 |
| 4 | 17 |
+---------+----------+
5 rows in set (0.00 sec)
音乐表：music，第一列是音乐 id，第二列是音乐 name，id 是主键，表中数据如下：
mysql> select * from music;
+----+------------+
| id | music_name |
+----+------------+
| 17 | yueyawang |
| 18 | kong |
| 19 | MOM |
| 20 | Sold Out |
+----+------------+
4 rows in set (0.02 sec)
【问题】请编写一个 SQL 语句，查询向 user_id = 1 的用户，推荐其关注的人喜欢的音乐。不要推荐该用户已经喜欢的音乐，并且按 music 的 id 升序排列。查询结果中不包含重复项。
查询结果如下：
| id | music_name |
|---|---|
| 18 | kong |
| 19 | MOM |
解答：
/* select distinct m.id, m.music_name from music m join (select user_id, music_id from music_likes where user_id in (select follower_id from follow where user_id = 1)) a on a.music_id = m.id where a.music_id not in (select music_id from music_likes where user_id = 1) order by m.id; */
mysql> select distinct m.id, m.music_name
-> from music m join
-> (select user_id, music_id from music_likes
-> where user_id in (select follower_id from follow where user_id = 1)) a
-> on a.music_id = m.id
-> where a.music_id not in (select music_id from music_likes where user_id = 1)
-> order by m.id;
+----+------------+
| id | music_name |
+----+------------+
| 18 | kong |
| 19 | MOM |
+----+------------+
2 rows in set (0.02 sec)
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