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数据库系统原理与应用教程（061）—— MySQL 练习题：操作题 21-31（五）

2022-07-28 12:30:00 【睿思达DBA_WGX】

数据库系统原理与应用教程（061）—— MySQL 练习题：操作题 21-31（五）

21、连接查询（1）

题目：要查看所有来自浙江大学的用户回答问题情况，取出相应数据。

示例：question_practice_detail 表的数据如下。

id	device_id	question_id	result
1	2138	111	wrong
2	3214	112	wrong
3	3214	113	wrong
4	6543	114	right
5	2315	115	right
6	2315	116	right
7	2315	117	wrong

示例：user_profile 表的数据如下。

id	device_id	gender	age	university	gpa	active_days_within_30	question_cnt	answer_cnt
1	2138	male	21	北京大学	3.4	7	2	12
2	3214	male		复旦大学	4.0	15	5	25
3	6543	female	20	北京大学	3.2	12	3	30
4	2315	female	23	浙江大学	3.6	5	1	2
5	5432	male	25	山东大学	3.8	20	15	70
6	2131	male	28	山东大学	3.3	15	7	13
7	4321	female	26	复旦大学	3.6	9	6	52

查询应返回以下结果，查询结果根据 question_id 升序排序：

device_id	question_id	result
2315	115	right
2315	116	right
2315	117	wrong

表结构及数据如下：

/* drop table if exists `user_profile`; drop table if exists `question_practice_detail`; CREATE TABLE `user_profile` ( `id` int NOT NULL, `device_id` int NOT NULL, `gender` varchar(14) NOT NULL, `age` int , `university` varchar(32) NOT NULL, `gpa` float, `active_days_within_30` int , `question_cnt` int , `answer_cnt` int ); CREATE TABLE `question_practice_detail` ( `id` int NOT NULL, `device_id` int NOT NULL, `question_id`int NOT NULL, `result` varchar(32) NOT NULL ); INSERT INTO user_profile VALUES(1,2138,'male',21,'北京大学',3.4,7,2,12); INSERT INTO user_profile VALUES(2,3214,'male',null,'复旦大学',4.0,15,5,25); INSERT INTO user_profile VALUES(3,6543,'female',20,'北京大学',3.2,12,3,30); INSERT INTO user_profile VALUES(4,2315,'female',23,'浙江大学',3.6,5,1,2); INSERT INTO user_profile VALUES(5,5432,'male',25,'山东大学',3.8,20,15,70); INSERT INTO user_profile VALUES(6,2131,'male',28,'山东大学',3.3,15,7,13); INSERT INTO user_profile VALUES(7,4321,'male',28,'复旦大学',3.6,9,6,52); INSERT INTO question_practice_detail VALUES(1,2138,111,'wrong'); INSERT INTO question_practice_detail VALUES(2,3214,112,'wrong'); INSERT INTO question_practice_detail VALUES(3,3214,113,'wrong'); INSERT INTO question_practice_detail VALUES(4,6543,111,'right'); INSERT INTO question_practice_detail VALUES(5,2315,115,'right'); INSERT INTO question_practice_detail VALUES(6,2315,116,'right'); INSERT INTO question_practice_detail VALUES(7,2315,117,'wrong'); */

解答：

/* -- 使用连接查询 select q.device_id, q.question_id, q.result from question_practice_detail q join user_profile u on q.device_id = u.device_id where u.university = '浙江大学'; -- 使用子查询 select device_id, question_id, result from question_practice_detail where device_id in (select device_id from user_profile where university = '浙江大学'); */
-- 使用连接查询
mysql> select q.device_id, q.question_id, q.result 
    -> from question_practice_detail q join user_profile u
    -> on q.device_id = u.device_id
    -> where u.university = '浙江大学';
+-----------+-------------+--------+
| device_id | question_id | result |
+-----------+-------------+--------+
|      2315 |         115 | right  |
|      2315 |         116 | right  |
|      2315 |         117 | wrong  |
+-----------+-------------+--------+
3 rows in set (0.00 sec)

-- 使用子查询
mysql> select device_id, question_id, result 
    -> from question_practice_detail 
    -> where device_id in 
    -> (select device_id from user_profile where university = '浙江大学');
+-----------+-------------+--------+
| device_id | question_id | result |
+-----------+-------------+--------+
|      2315 |         115 | right  |
|      2315 |         116 | right  |
|      2315 |         117 | wrong  |
+-----------+-------------+--------+
3 rows in set (0.01 sec)

22、连接查询（2）

题目：查询每个学校答过题的用户平均答题数量，取出相关数据。

示例：user_profile 表的数据如下。

device_id	gender	age	university	gpa	active_days_within_30
2138	male	21	北京大学	3.4	7
3214	male	NULL	复旦大学	4	15
6543	female	20	北京大学	3.2	12
2315	female	23	浙江大学	3.6	5
5432	male	25	山东大学	3.8	20
2131	male	28	山东大学	3.3	15
4321	male	28	复旦大学	3.6	9

示例：question_practice_detail 表的数据如下。

device_id	question_id	result
2138	111	wrong
3214	112	wrong
3214	113	wrong
6543	111	right
2315	115	right
2315	116	right
2315	117	wrong
5432	118	wrong
5432	112	wrong
2131	114	right
5432	113	wrong

查找每个学校用户的平均答题数目（说明：某学校用户平均答题数量计算方式为该学校用户答题总次数除以答过题的不同用户个数），查询应返回以下结果（结果保留4位小数），按照 university 升序排序。

university	avg_answer_cnt
北京大学	1.0000
复旦大学	2.0000
山东大学	2.0000
浙江大学	3.0000

表结构及数据如下：

/* drop table if exists `user_profile`; drop table if exists `question_practice_detail`; CREATE TABLE `user_profile` ( `device_id` int NOT NULL, `gender` varchar(14) NOT NULL, `age` int , `university` varchar(32) NOT NULL, `gpa` float, `active_days_within_30` int ); CREATE TABLE `question_practice_detail` ( `device_id` int NOT NULL, `question_id`int NOT NULL, `result` varchar(32) NOT NULL ); INSERT INTO user_profile VALUES(2138,'male',21,'北京大学',3.4,7); INSERT INTO user_profile VALUES(3214,'male',null,'复旦大学',4.0,15); INSERT INTO user_profile VALUES(6543,'female',20,'北京大学',3.2,12); INSERT INTO user_profile VALUES(2315,'female',23,'浙江大学',3.6,5); INSERT INTO user_profile VALUES(5432,'male',25,'山东大学',3.8,20); INSERT INTO user_profile VALUES(2131,'male',28,'山东大学',3.3,15); INSERT INTO user_profile VALUES(4321,'male',28,'复旦大学',3.6,9); INSERT INTO question_practice_detail VALUES(2138,111,'wrong'); INSERT INTO question_practice_detail VALUES(3214,112,'wrong'); INSERT INTO question_practice_detail VALUES(3214,113,'wrong'); INSERT INTO question_practice_detail VALUES(6543,111,'right'); INSERT INTO question_practice_detail VALUES(2315,115,'right'); INSERT INTO question_practice_detail VALUES(2315,116,'right'); INSERT INTO question_practice_detail VALUES(2315,117,'wrong'); INSERT INTO question_practice_detail VALUES(5432,118,'wrong'); INSERT INTO question_practice_detail VALUES(5432,112,'wrong'); INSERT INTO question_practice_detail VALUES(2131,114,'right'); INSERT INTO question_practice_detail VALUES(5432,113,'wrong'); */

解答：

/* select u.university, round(count(*)/count(distinct u.device_id),4) avg_answer_cnt from question_practice_detail q join user_profile u on q.device_id = u.device_id group by u.university order by u.university; */
mysql> select u.university, 
    ->        round(count(*)/count(distinct u.device_id),4) avg_answer_cnt
    -> from question_practice_detail q join user_profile u
    -> on q.device_id = u.device_id
    -> group by u.university
    -> order by u.university;
+--------------+----------------+
| university   | avg_answer_cnt |
+--------------+----------------+
| 北京大学     |         1.0000 |
| 复旦大学     |         2.0000 |
| 山东大学     |         2.0000 |
| 浙江大学     |         3.0000 |
+--------------+----------------+
4 rows in set (0.00 sec)

23、连接查询（3）

题目：查询参加了答题的不同学校、不同难度的用户平均答题量，取出相应数据。

示例：user_profile（用户信息表）的数据如下。

id	device_id	gender	age	university	gpa	active_days_within_30	question_cnt	answer_cnt
1	2138	male	21	北京大学	3.4	7	2	12
2	3214	male	NULL	复旦大学	4	15	5	25
3	6543	female	20	北京大学	3.2	12	3	30
4	2315	female	23	浙江大学	3.6	5	1	2
5	5432	male	25	山东大学	3.8	20	15	70
6	2131	male	28	山东大学	3.3	15	7	13
7	4321	male	28	复旦大学	3.6	9	6	52

示例：question_practice_detail（题库练习明细表）的数据如下。

id	device_id	question_id	result
1	2138	111	wrong
2	3214	112	wrong
3	3214	113	wrong
4	6534	111	right
5	2315	115	right
6	2315	116	right
7	2315	117	wrong
8	5432	117	wrong
9	5432	112	wrong
10	2131	113	right
11	5432	113	wrong
12	2315	115	right
13	2315	116	right
14	2315	117	wrong
15	5432	117	wrong
16	5432	112	wrong
17	2131	113	right
18	5432	113	wrong
19	2315	117	wrong
20	5432	117	wrong
21	5432	112	wrong
22	2131	113	right
23	5432	113	wrong

示例：question_detail（用户信息表）的数据如下。

id	question_id	difficult_level
1	111	hard
2	112	medium
3	113	easy
4	115	easy
5	116	medium
6	117	easy

请写出 SQL 语句，计算不同学校、不同难度的用户平均答题量，查询应返回以下结果（数据四舍五入保留 4 位小数)：

university	difficult_level	avg_answer_cnt
北京大学	hard	1.0000
复旦大学	easy	1.0000
复旦大学	medium	1.0000
山东大学	easy	4.5000
山东大学	medium	3.0000
浙江大学	easy	5.0000
浙江大学	medium	2.0000

表结构及数据如下：

/* drop table if exists `user_profile`; drop table if exists `question_practice_detail`; drop table if exists `question_detail`; CREATE TABLE `user_profile` ( `id` int NOT NULL, `device_id` int NOT NULL, `gender` varchar(14) NOT NULL, `age` int , `university` varchar(32) NOT NULL, `gpa` float, `active_days_within_30` int , `question_cnt` int , `answer_cnt` int ); CREATE TABLE `question_practice_detail` ( `id` int NOT NULL, `device_id` int NOT NULL, `question_id`int NOT NULL, `result` varchar(32) NOT NULL ); CREATE TABLE `question_detail` ( `id` int NOT NULL, `question_id`int NOT NULL, `difficult_level` varchar(32) NOT NULL ); INSERT INTO user_profile VALUES(1,2138,'male',21,'北京大学',3.4,7,2,12); INSERT INTO user_profile VALUES(2,3214,'male',null,'复旦大学',4.0,15,5,25); INSERT INTO user_profile VALUES(3,6543,'female',20,'北京大学',3.2,12,3,30); INSERT INTO user_profile VALUES(4,2315,'female',23,'浙江大学',3.6,5,1,2); INSERT INTO user_profile VALUES(5,5432,'male',25,'山东大学',3.8,20,15,70); INSERT INTO user_profile VALUES(6,2131,'male',28,'山东大学',3.3,15,7,13); INSERT INTO user_profile VALUES(7,4321,'male',28,'复旦大学',3.6,9,6,52); INSERT INTO question_practice_detail VALUES(1,2138,111,'wrong'); INSERT INTO question_practice_detail VALUES(2,3214,112,'wrong'); INSERT INTO question_practice_detail VALUES(3,3214,113,'wrong'); INSERT INTO question_practice_detail VALUES(4,6543,111,'right'); INSERT INTO question_practice_detail VALUES(5,2315,115,'right'); INSERT INTO question_practice_detail VALUES(6,2315,116,'right'); INSERT INTO question_practice_detail VALUES(7,2315,117,'wrong'); INSERT INTO question_practice_detail VALUES(8,5432,117,'wrong'); INSERT INTO question_practice_detail VALUES(9,5432,112,'wrong'); INSERT INTO question_practice_detail VALUES(10,2131,113,'right'); INSERT INTO question_practice_detail VALUES(11,5432,113,'wrong'); INSERT INTO question_practice_detail VALUES(12,2315,115,'right'); INSERT INTO question_practice_detail VALUES(13,2315,116,'right'); INSERT INTO question_practice_detail VALUES(14,2315,117,'wrong'); INSERT INTO question_practice_detail VALUES(15,5432,117,'wrong'); INSERT INTO question_practice_detail VALUES(16,5432,112,'wrong'); INSERT INTO question_practice_detail VALUES(17,2131,113,'right'); INSERT INTO question_practice_detail VALUES(18,5432,113,'wrong'); INSERT INTO question_practice_detail VALUES(19,2315,117,'wrong'); INSERT INTO question_practice_detail VALUES(20,5432,117,'wrong'); INSERT INTO question_practice_detail VALUES(21,5432,112,'wrong'); INSERT INTO question_practice_detail VALUES(22,2131,113,'right'); INSERT INTO question_practice_detail VALUES(23,5432,113,'wrong'); INSERT INTO question_detail VALUES(1,111,'hard'); INSERT INTO question_detail VALUES(2,112,'medium'); INSERT INTO question_detail VALUES(3,113,'easy'); INSERT INTO question_detail VALUES(4,115,'easy'); INSERT INTO question_detail VALUES(5,116,'medium'); INSERT INTO question_detail VALUES(6,117,'easy'); */

解答：

/* select u.university, qd.difficult_level, round(count(*)/count(distinct u.device_id),4) avg_answer_cnt from user_profile u join question_practice_detail q on u.device_id = q.device_id join question_detail qd on q.question_id = qd.question_id group by u.university, qd.difficult_level; */
mysql> select u.university, qd.difficult_level,
    ->        round(count(*)/count(distinct u.device_id),4) avg_answer_cnt
    -> from user_profile u join question_practice_detail q
    ->      on u.device_id = q.device_id
    ->      join question_detail qd on q.question_id = qd.question_id
    -> group by u.university, qd.difficult_level;
+--------------+-----------------+----------------+
| university   | difficult_level | avg_answer_cnt |
+--------------+-----------------+----------------+
| 北京大学     | hard            |         1.0000 |
| 复旦大学     | easy            |         1.0000 |
| 复旦大学     | medium          |         1.0000 |
| 山东大学     | easy            |         4.5000 |
| 山东大学     | medium          |         3.0000 |
| 浙江大学     | easy            |         5.0000 |
| 浙江大学     | medium          |         2.0000 |
+--------------+-----------------+----------------+
7 rows in set (0.00 sec)

24、连接查询（4）

题目：查询参加了答题的山东大学的用户在不同难度下的平均答题题目数，取出相应数据。

示例：user_profile（用户信息表）的数据如下。

id	device_id	gender	age	university	gpa	active_days_within_30	question_cnt	answer_cnt
1	2138	male	21	北京大学	3.4	7	2	12
2	3214	male	NULL	复旦大学	4	15	5	25
3	6543	female	20	北京大学	3.2	12	3	30
4	2315	female	23	浙江大学	3.6	5	1	2
5	5432	male	25	山东大学	3.8	20	15	70
6	2131	male	28	山东大学	3.3	15	7	13
7	4321	male	28	复旦大学	3.6	9	6	52

示例：question_practice_detail（题库练习明细表）的数据如下。

id	device_id	question_id	result
1	2138	111	wrong
2	3214	112	wrong
3	3214	113	wrong
4	6534	111	right
5	2315	115	right
6	2315	116	right
7	2315	117	wrong
8	5432	117	wrong
9	5432	112	wrong
10	2131	113	right
11	5432	113	wrong
12	2315	115	right
13	2315	116	right
14	2315	117	wrong
15	5432	117	wrong
16	5432	112	wrong
17	2131	113	right
18	5432	113	wrong
19	2315	117	wrong
20	5432	117	wrong
21	5432	112	wrong
22	2131	113	right
23	5432	113	wrong

示例：question_detail（用户信息表）的数据如下。

id	question_id	difficult_level
1	111	hard
2	112	medium
3	113	easy
4	115	easy
5	116	medium
6	117	easy

查询应返回以下结果（数据四舍五入保留 4 位小数）：

university	difficult_level	avg_answer_cnt
山东大学	easy	4.5000
山东大学	medium	3.0000

解答：

/* select u.university, qd.difficult_level, round(count(*)/count(distinct u.device_id),4) avg_answer_cnt from user_profile u join question_practice_detail q on u.device_id = q.device_id join question_detail qd on q.question_id = qd.question_id where u.university = '山东大学' group by u.university, qd.difficult_level; */
mysql> select u.university, qd.difficult_level,
    ->        round(count(*)/count(distinct u.device_id),4) avg_answer_cnt
    -> from user_profile u join question_practice_detail q
    ->      on u.device_id = q.device_id
    ->      join question_detail qd on q.question_id = qd.question_id
    -> where u.university = '山东大学'
    -> group by u.university, qd.difficult_level;
+--------------+-----------------+----------------+
| university   | difficult_level | avg_answer_cnt |
+--------------+-----------------+----------------+
| 山东大学     | easy            |         4.5000 |
| 山东大学     | medium          |         3.0000 |
+--------------+-----------------+----------------+
2 rows in set (0.00 sec)

25、联合查询（union）

题目：查看学校为山东大学或者性别为男性的用户的 device_id、gender、age 和 gpa数据，取出相应结果（结果不去重）。

示例：user_profile 的数据如下。

id	device_id	gender	age	university	gpa	active_days_within_30	question_cnt	answer_cnt
1	2138	male	21	北京大学	3.4	7	2	12
2	3214	male		复旦大学	4	15	5	25
3	6543	female	20	北京大学	3.2	12	3	30
4	2315	female	23	浙江大学	3.6	5	1	2
5	5432	male	25	山东大学	3.8	20	15	70
6	2131	male	28	山东大学	3.3	15	7	13
7	4321	male	26	复旦大学	3.6	9	6	52

查询应返回以下结果：

device_id	gender	age	gpa
5432	male	25	3.8
2131	male	28	3.3
2138	male	21	3.4
3214	male	None	4
5432	male	25	3.8
2131	male	28	3.3
4321	male	28	3.6

解答：

/* select device_id, gender, age, gpa from user_profile where university= '山东大学' union all select device_id, gender, age, gpa from user_profile where gender = 'male'; */
mysql> select device_id, gender, age, gpa from user_profile where university= '山东大学'
    -> union all
    -> select device_id, gender, age, gpa from user_profile where gender = 'male';
+-----------+--------+------+------+
| device_id | gender | age  | gpa  |
+-----------+--------+------+------+
|      5432 | male   |   25 |  3.8 |
|      2131 | male   |   28 |  3.3 |
|      2138 | male   |   21 |  3.4 |
|      3214 | male   | NULL |    4 |
|      5432 | male   |   25 |  3.8 |
|      2131 | male   |   28 |  3.3 |
|      4321 | male   |   28 |  3.6 |
+-----------+--------+------+------+
7 rows in set (0.00 sec)

26、IF 函数的使用

题目：将用户划分为 25 岁以下和 25 岁及以上两个年龄段，分别查询这两个年龄段用户数量（age为null 也记为 25岁以下）。

示例：user_profile 的数据如下。

id	device_id	gender	age	university	gpa	active_days_within_30	question_cnt	answer_cnt
1	2138	male	21	北京大学	3.4	7	2	12
2	3214	male		复旦大学	4	15	5	25
3	6543	female	20	北京大学	3.2	12	3	30
4	2315	female	23	浙江大学	3.6	5	1	2
5	5432	male	25	山东大学	3.8	20	15	70
6	2131	male	28	山东大学	3.3	15	7	13
7	4321	male	26	复旦大学	3.6	9	6	52

查询应返回以下结果：

age_cut	number
25 岁以下	4
25 岁及以上	3

/* select '25 岁以下' age_cut,sum(if(age >=25, 0, 1)) number from user_profile union select '25 岁及以上' age_cut,sum(if(age >=25, 1, 0)) number from user_profile; select '25 岁以下' age_cut, count(*) number from user_profile where age < 25 or age is null union select '25 岁及以上' age_cut, count(*) number from user_profile where age >=25; select if(age >=25, '25 岁及以上', '25 岁以下') age_cut, count(*) number from user_profile group by age_cut; */
mysql> select if(age >=25, '25 岁及以上', '25 岁以下') age_cut, count(*) number 
    -> from user_profile group by age_cut; 
+-----------------+--------+
| age_cut         | number |
+-----------------+--------+
| 25 岁以下       |      4 |
| 25 岁及以上     |      3 |
+-----------------+--------+
2 rows in set (0.00 sec)

mysql> select '25 岁以下' age_cut,sum(if(age >=25, 0, 1)) number from user_profile
    -> union
    -> select '25 岁及以上' age_cut,sum(if(age >=25, 1, 0)) number from user_profile;
+-----------------+--------+
| age_cut         | number |
+-----------------+--------+
| 25 岁以下       |      4 |
| 25 岁及以上     |      3 |
+-----------------+--------+
2 rows in set (0.00 sec)

mysql> select '25 岁以下' age_cut, count(*) number from user_profile where age < 25 or age is null
    -> union
    -> select '25 岁及以上' age_cut, count(*) number from user_profile where age >=25;
+-----------------+--------+
| age_cut         | number |
+-----------------+--------+
| 25 岁以下       |      4 |
| 25 岁及以上     |      3 |
+-----------------+--------+
2 rows in set (0.00 sec)

27、CASE 函数的使用

题目：将用户划分为 20 岁以下，20-24岁，25岁及以上三个年龄段，分别查看不同年龄段用户的明细情况，取出相应数据。（注：若年龄为空则返回其他）

示例：user_profile 的数据如下。

id	device_id	gender	age	university	gpa	active_days_within_30	question_cnt	answer_cnt
1	2138	male	21	北京大学	3.4	7	2	12
2	3214	male		复旦大学	4	15	5	25
3	6543	female	20	北京大学	3.2	12	3	30
4	2315	female	23	浙江大学	3.6	5	1	2
5	5432	male	25	山东大学	3.8	20	15	70
6	2131	male	28	山东大学	3.3	15	7	13
7	4321	male	26	复旦大学	3.6	9	6	52

查询应返回以下结果：

device_id	gender	age_cut
2138	male	20-24岁
3214	male	其他
6543	female	20-24岁
2315	female	20-24岁
5432	male	25岁及以上
2131	male	25岁及以上
4321	male	25岁及以上

解答：

/* select device_id, gender, case when age is null then '其他' when age >=20 and age <= 24 then '20-24岁' when age > 24 then '25岁及以上' end age_cut from user_profile; */

mysql> select device_id, gender, 
    ->        case 
    ->            when age is null then '其他' 
    ->            when age >=20 and age <= 24 then '20-24岁' 
    ->            when age > 24 then '25岁及以上' 
    ->         end age_cut 
    -> from user_profile;
+-----------+--------+----------------+
| device_id | gender | age_cut        |
+-----------+--------+----------------+
|      2138 | male   | 20-24岁        |
|      3214 | male   | 其他           |
|      6543 | female | 20-24岁        |
|      2315 | female | 20-24岁        |
|      5432 | male   | 25岁及以上     |
|      2131 | male   | 25岁及以上     |
|      4321 | male   | 25岁及以上     |
+-----------+--------+----------------+
7 rows in set (0.01 sec)

28、日期函数（YEAR）的使用

题目：计算出 2021 年 8 月每天用户练习题目的数量，取出相应数据。

示例：question_practice_detail 的数据如下。

id	device_id	question_id	result	date
1	2138	111	wrong	2021-05-03
2	3214	112	wrong	2021-05-09
3	3214	113	wrong	2021-06-15
4	6543	111	right	2021-08-13
5	2315	115	right	2021-08-13
6	2315	116	right	2021-08-14
7	2315	117	wrong	2021-08-15
……

查询应返回以下结果：

day	question_cnt
13	5
14	2
15	3
16	1
18	1

表结构及数据如下：

/* CREATE TABLE `question_practice_detail` ( `id` int NOT NULL, `device_id` int NOT NULL, `question_id`int NOT NULL, `result` varchar(32) NOT NULL, `date` date NOT NULL ); INSERT INTO question_practice_detail VALUES(1,2138,111,'wrong','2021-05-03'); INSERT INTO question_practice_detail VALUES(2,3214,112,'wrong','2021-05-09'); INSERT INTO question_practice_detail VALUES(3,3214,113,'wrong','2021-06-15'); INSERT INTO question_practice_detail VALUES(4,6543,111,'right','2021-08-13'); INSERT INTO question_practice_detail VALUES(5,2315,115,'right','2021-08-13'); INSERT INTO question_practice_detail VALUES(6,2315,116,'right','2021-08-14'); INSERT INTO question_practice_detail VALUES(7,2315,117,'wrong','2021-08-15'); INSERT INTO question_practice_detail VALUES(8,3214,112,'wrong','2021-05-09'); INSERT INTO question_practice_detail VALUES(9,3214,113,'wrong','2021-08-15'); INSERT INTO question_practice_detail VALUES(10,6543,111,'right','2021-08-13'); INSERT INTO question_practice_detail VALUES(11,2315,115,'right','2021-08-13'); INSERT INTO question_practice_detail VALUES(12,2315,116,'right','2021-08-14'); INSERT INTO question_practice_detail VALUES(13,2315,117,'wrong','2021-08-15'); INSERT INTO question_practice_detail VALUES(14,3214,112,'wrong','2021-08-16'); INSERT INTO question_practice_detail VALUES(15,3214,113,'wrong','2021-08-18'); INSERT INTO question_practice_detail VALUES(16,6543,111,'right','2021-08-13'); */

解答：

/* select day(date) day, count(*) question_cnt from question_practice_detail where date between '2021-8-1' and '2021-8-31' group by day; */
mysql> select day(date) day, count(*) question_cnt
    -> from question_practice_detail 
    -> where date between '2021-8-1' and '2021-8-31'
    -> group by day;
+------+--------------+
| day  | question_cnt |
+------+--------------+
|   13 |            5 |
|   14 |            2 |
|   15 |            3 |
|   16 |            1 |
|   18 |            1 |
+------+--------------+
5 rows in set (0.00 sec)

29、SUBSTRING_INDEX 函数的使用（1）

题目：统计每个性别的用户分别有多少参赛者，取出相应结果。

示例：user_submit 的数据如下。

device_id	profile	blog_url
2138	180cm,75kg,27,male	http:/url/bigboy777
3214	165cm,45kg,26,female	http:/url/kittycc
6543	178cm,65kg,25,male	http:/url/tiger
4321	171cm,55kg,23,female	http:/url/uhksd
2131	168cm,45kg,22,female	http:/urlsydney

查询应返回以下结果：

gender	number
male	2
female	3

表结构及数据如下：

/* drop table if exists user_submit; CREATE TABLE `user_submit` ( `id` int NOT NULL, `device_id` int NOT NULL, `profile` varchar(100) NOT NULL, `blog_url` varchar(100) NOT NULL ); INSERT INTO user_submit VALUES(1,2138,'180cm,75kg,27,male','http:/url/bisdgboy777'); INSERT INTO user_submit VALUES(1,3214,'165cm,45kg,26,female','http:/url/dkittycc'); INSERT INTO user_submit VALUES(1,6543,'178cm,65kg,25,male','http:/url/tigaer'); INSERT INTO user_submit VALUES(1,4321,'171cm,55kg,23,female','http:/url/uhsksd'); INSERT INTO user_submit VALUES(1,2131,'168cm,45kg,22,female','http:/url/sysdney'); */

解答：

/* select SUBSTRING_INDEX(profile,',',-1) gender, count(*) number from user_submit group by gender; */

mysql> select SUBSTRING_INDEX(profile,',',-1) gender, count(*) number
    -> from user_submit
    -> group by gender;
+--------+--------+
| gender | number |
+--------+--------+
| female |      3 |
| male   |      2 |
+--------+--------+
2 rows in set (0.02 sec)

30、SUBSTRING_INDEX 函数的使用（2）

题目：blog_url 列中 url 字符后的字符串为用户个人博客的用户名，取出用户名作为一个新列，取出所需数据。

示例：user_submit 的数据如下。

device_id	profile	blog_url
2138	180cm,75kg,27,male	http:/ur/bisdgboy777
3214	165cm,45kg,26,female	http:/url/dkittycc
6543	178cm,65kg,25,male	http:/ur/tigaer
4321	171 cm,55kg,23,female	http:/url/uhksd
2131	168cm,45kg,22,female	http:/url/sydney

查询应返回以下结果：

device_id	user_name
2138	bisdgboy777
3214	dkittycc
6543	tigaer
4321	uhsksd
2131	sydney

解答：

/* select device_id, SUBSTRING_INDEX(blog_url,'/',-1) user_name from user_submit; */
mysql> select device_id,
    ->        SUBSTRING_INDEX(blog_url,'/',-1) user_name
    -> from user_submit;
+-----------+-------------+
| device_id | user_name   |
+-----------+-------------+
|      2138 | bisdgboy777 |
|      3214 | dkittycc    |
|      6543 | tigaer      |
|      4321 | uhsksd      |
|      2131 | sysdney     |
+-----------+-------------+
5 rows in set (0.00 sec)

31、SUBSTRING_INDEX 函数的使用（3）

题目：统计每个年龄的用户分别有多少参赛者，取出相应结果。

示例：user_submit 的数据如下。

device_id	profile	blog_url
2138	180cm,75kg,27,male	http:/ur/bigboy777
3214	165cm,45kg,26,female	http:/url/kittycc
6543	178cm,65kg,25,male	http:/url/tiger
4321	171cm,55kg,23,female	http:/url/uhksd
2131	168cm,45kg,22,female	http:/url/sydney

查询应返回以下结果：

age	number
27	1
26	1
25	1
23	1
22	1

解答：

/* select SUBSTRING_INDEX(SUBSTRING_INDEX(profile,',',3),',',-1) age, count(*) number from user_submit group by age; */
mysql> select SUBSTRING_INDEX(SUBSTRING_INDEX(profile,',',3),',',-1) age,
    ->        count(*) number
    -> from user_submit
    -> group by age;
+------+--------+
| age  | number |
+------+--------+
| 22   |      1 |
| 23   |      1 |
| 25   |      1 |
| 26   |      1 |
| 27   |      1 |
+------+--------+
5 rows in set (0.00 sec)

原网站

版权声明
本文为[睿思达DBA_WGX]所创，转载请带上原文链接，感谢
https://wanggx.blog.csdn.net/article/details/126027811

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数据库系统原理与应用教程（061）—— MySQL 练习题：操作题 21-31（五）

数据库系统原理与应用教程（061）—— MySQL 练习题：操作题 21-31（五）

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