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Tutorial on the principle and application of database system (061) -- MySQL exercise: operation questions 21-31 (V)

2022-07-28 13:45:00 【Rsda DBA_ WGX】

Database system principle and Application Tutorial （061）—— MySQL Exercises ： Operation questions 21-31（ 5、 ... and ）

21、 Link query （1）

subject ： To view the answers of all users from Zhejiang University , Take out the corresponding data .

Example ：question_practice_detail The data in the table are as follows .

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

Example ：user_profile The data in the table are as follows .

id	device_id	gender	age	university	gpa	active_days_within_30	question_cnt	answer_cnt
1	2138	male	21	Peking University,	3.4	7	2	12
2	3214	male		Fudan University	4.0	15	5	25
3	6543	female	20	Peking University,	3.2	12	3	30
4	2315	female	23	Zhejiang University	3.6	5	1	2
5	5432	male	25	Shandong University	3.8	20	15	70
6	2131	male	28	Shandong University	3.3	15	7	13
7	4321	female	26	Fudan University	3.6	9	6	52

The query should return the following results , The query results are based on question_id Ascending sort ：

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

Table structure and data are as follows ：

/* 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,' Peking University, ',3.4,7,2,12); INSERT INTO user_profile VALUES(2,3214,'male',null,' Fudan University ',4.0,15,5,25); INSERT INTO user_profile VALUES(3,6543,'female',20,' Peking University, ',3.2,12,3,30); INSERT INTO user_profile VALUES(4,2315,'female',23,' Zhejiang University ',3.6,5,1,2); INSERT INTO user_profile VALUES(5,5432,'male',25,' Shandong University ',3.8,20,15,70); INSERT INTO user_profile VALUES(6,2131,'male',28,' Shandong University ',3.3,15,7,13); INSERT INTO user_profile VALUES(7,4321,'male',28,' Fudan University ',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'); */

answer ：

/* --  Use connection query  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 = ' Zhejiang University '; --  Use subquery  select device_id, question_id, result from question_practice_detail where device_id in (select device_id from user_profile where university = ' Zhejiang University '); */
--  Use connection query 
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 = ' Zhejiang University ';
+-----------+-------------+--------+
| device_id | question_id | result |
+-----------+-------------+--------+
|      2315 |         115 | right  |
|      2315 |         116 | right  |
|      2315 |         117 | wrong  |
+-----------+-------------+--------+
3 rows in set (0.00 sec)

--  Use subquery 
mysql> select device_id, question_id, result 
    -> from question_practice_detail 
    -> where device_id in 
    -> (select device_id from user_profile where university = ' Zhejiang University ');
+-----------+-------------+--------+
| device_id | question_id | result |
+-----------+-------------+--------+
|      2315 |         115 | right  |
|      2315 |         116 | right  |
|      2315 |         117 | wrong  |
+-----------+-------------+--------+
3 rows in set (0.01 sec)

22、 Link query （2）

subject ： Query the average number of users who have answered questions in each school , Take out relevant data .

Example ：user_profile The data in the table are as follows .

device_id	gender	age	university	gpa	active_days_within_30
2138	male	21	Peking University,	3.4	7
3214	male	NULL	Fudan University	4	15
6543	female	20	Peking University,	3.2	12
2315	female	23	Zhejiang University	3.6	5
5432	male	25	Shandong University	3.8	20
2131	male	28	Shandong University	3.3	15
4321	male	28	Fudan University	3.6	9

Example ：question_practice_detail The data in the table are as follows .

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

Find the average number of answers per school user （ explain ： The average number of questions answered by users in a school is calculated by dividing the total number of answers by the number of different users who have answered the questions ）, The query should return the following results （ The result is reserved 4 Decimal place ）, according to university Ascending sort .

university	avg_answer_cnt
Peking University,	1.0000
Fudan University	2.0000
Shandong University	2.0000
Zhejiang University	3.0000

Table structure and data are as follows ：

/* 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,' Peking University, ',3.4,7); INSERT INTO user_profile VALUES(3214,'male',null,' Fudan University ',4.0,15); INSERT INTO user_profile VALUES(6543,'female',20,' Peking University, ',3.2,12); INSERT INTO user_profile VALUES(2315,'female',23,' Zhejiang University ',3.6,5); INSERT INTO user_profile VALUES(5432,'male',25,' Shandong University ',3.8,20); INSERT INTO user_profile VALUES(2131,'male',28,' Shandong University ',3.3,15); INSERT INTO user_profile VALUES(4321,'male',28,' Fudan University ',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'); */

answer ：

/* 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 |
+--------------+----------------+
|  Peking University,      |         1.0000 |
|  Fudan University      |         2.0000 |
|  Shandong University      |         2.0000 |
|  Zhejiang University      |         3.0000 |
+--------------+----------------+
4 rows in set (0.00 sec)

23、 Link query （3）

subject ： Check the different schools that participated in the answer 、 The average number of questions answered by users with different difficulties , Take out the corresponding data .

Example ：user_profile（ User information sheet ） The data are as follows .

id	device_id	gender	age	university	gpa	active_days_within_30	question_cnt	answer_cnt
1	2138	male	21	Peking University,	3.4	7	2	12
2	3214	male	NULL	Fudan University	4	15	5	25
3	6543	female	20	Peking University,	3.2	12	3	30
4	2315	female	23	Zhejiang University	3.6	5	1	2
5	5432	male	25	Shandong University	3.8	20	15	70
6	2131	male	28	Shandong University	3.3	15	7	13
7	4321	male	28	Fudan University	3.6	9	6	52

Example ：question_practice_detail（ List of exercises in question bank ） The data are as follows .

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

Example ：question_detail（ User information sheet ） The data are as follows .

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

Please write out SQL sentence , Calculate different schools 、 The average number of questions answered by users with different difficulties , The query should return the following results （ Data rounding is reserved 4 Decimal place )：

university	difficult_level	avg_answer_cnt
Peking University,	hard	1.0000
Fudan University	easy	1.0000
Fudan University	medium	1.0000
Shandong University	easy	4.5000
Shandong University	medium	3.0000
Zhejiang University	easy	5.0000
Zhejiang University	medium	2.0000

Table structure and data are as follows ：

/* 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,' Peking University, ',3.4,7,2,12); INSERT INTO user_profile VALUES(2,3214,'male',null,' Fudan University ',4.0,15,5,25); INSERT INTO user_profile VALUES(3,6543,'female',20,' Peking University, ',3.2,12,3,30); INSERT INTO user_profile VALUES(4,2315,'female',23,' Zhejiang University ',3.6,5,1,2); INSERT INTO user_profile VALUES(5,5432,'male',25,' Shandong University ',3.8,20,15,70); INSERT INTO user_profile VALUES(6,2131,'male',28,' Shandong University ',3.3,15,7,13); INSERT INTO user_profile VALUES(7,4321,'male',28,' Fudan University ',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'); */

answer ：

/* 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 |
+--------------+-----------------+----------------+
|  Peking University,      | hard            |         1.0000 |
|  Fudan University      | easy            |         1.0000 |
|  Fudan University      | medium          |         1.0000 |
|  Shandong University      | easy            |         4.5000 |
|  Shandong University      | medium          |         3.0000 |
|  Zhejiang University      | easy            |         5.0000 |
|  Zhejiang University      | medium          |         2.0000 |
+--------------+-----------------+----------------+
7 rows in set (0.00 sec)

24、 Link query （4）

subject ： Query the average number of questions answered by users of Shandong University who participated in the answer under different difficulties , Take out the corresponding data .

Example ：user_profile（ User information sheet ） The data are as follows .

id	device_id	gender	age	university	gpa	active_days_within_30	question_cnt	answer_cnt
1	2138	male	21	Peking University,	3.4	7	2	12
2	3214	male	NULL	Fudan University	4	15	5	25
3	6543	female	20	Peking University,	3.2	12	3	30
4	2315	female	23	Zhejiang University	3.6	5	1	2
5	5432	male	25	Shandong University	3.8	20	15	70
6	2131	male	28	Shandong University	3.3	15	7	13
7	4321	male	28	Fudan University	3.6	9	6	52

Example ：question_practice_detail（ List of exercises in question bank ） The data are as follows .

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

Example ：question_detail（ User information sheet ） The data are as follows .

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

The query should return the following results （ Data rounding is reserved 4 Decimal place ）：

university	difficult_level	avg_answer_cnt
Shandong University	easy	4.5000
Shandong University	medium	3.0000

answer ：

/* 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 = ' Shandong University ' 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 = ' Shandong University '
    -> group by u.university, qd.difficult_level;
+--------------+-----------------+----------------+
| university   | difficult_level | avg_answer_cnt |
+--------------+-----------------+----------------+
|  Shandong University      | easy            |         4.5000 |
|  Shandong University      | medium          |         3.0000 |
+--------------+-----------------+----------------+
2 rows in set (0.00 sec)

25、 The joint query （union）

subject ： View users whose school is Shandong University or whose gender is male device_id、gender、age and gpa data , Take out the corresponding results （ The result is no weight loss ）.

Example ：user_profile The data are as follows .

id	device_id	gender	age	university	gpa	active_days_within_30	question_cnt	answer_cnt
1	2138	male	21	Peking University,	3.4	7	2	12
2	3214	male		Fudan University	4	15	5	25
3	6543	female	20	Peking University,	3.2	12	3	30
4	2315	female	23	Zhejiang University	3.6	5	1	2
5	5432	male	25	Shandong University	3.8	20	15	70
6	2131	male	28	Shandong University	3.3	15	7	13
7	4321	male	26	Fudan University	3.6	9	6	52

The query should return the following results ：

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

answer ：

/* select device_id, gender, age, gpa from user_profile where university= ' Shandong University ' 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= ' Shandong University '
    -> 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 Use of functions

subject ： Divide users into 25 Under and 25 Two age groups aged and over , Query the number of users in these two age groups （age by null It's also recorded as 25 Under the age of ）.

Example ：user_profile The data are as follows .

id	device_id	gender	age	university	gpa	active_days_within_30	question_cnt	answer_cnt
1	2138	male	21	Peking University,	3.4	7	2	12
2	3214	male		Fudan University	4	15	5	25
3	6543	female	20	Peking University,	3.2	12	3	30
4	2315	female	23	Zhejiang University	3.6	5	1	2
5	5432	male	25	Shandong University	3.8	20	15	70
6	2131	male	28	Shandong University	3.3	15	7	13
7	4321	male	26	Fudan University	3.6	9	6	52

The query should return the following results ：

age_cut	number
25 Under the age of	4
25 Years old and over	3

/* select '25  Under the age of ' age_cut,sum(if(age >=25, 0, 1)) number from user_profile union select '25  Years old and over ' age_cut,sum(if(age >=25, 1, 0)) number from user_profile; select '25  Under the age of ' age_cut, count(*) number from user_profile where age < 25 or age is null union select '25  Years old and over ' age_cut, count(*) number from user_profile where age >=25; select if(age >=25, '25  Years old and over ', '25  Under the age of ') age_cut, count(*) number from user_profile group by age_cut; */
mysql> select if(age >=25, '25  Years old and over ', '25  Under the age of ') age_cut, count(*) number 
    -> from user_profile group by age_cut; 
+-----------------+--------+
| age_cut         | number |
+-----------------+--------+
| 25  Under the age of        |      4 |
| 25  Years old and over      |      3 |
+-----------------+--------+
2 rows in set (0.00 sec)

mysql> select '25  Under the age of ' age_cut,sum(if(age >=25, 0, 1)) number from user_profile
    -> union
    -> select '25  Years old and over ' age_cut,sum(if(age >=25, 1, 0)) number from user_profile;
+-----------------+--------+
| age_cut         | number |
+-----------------+--------+
| 25  Under the age of        |      4 |
| 25  Years old and over      |      3 |
+-----------------+--------+
2 rows in set (0.00 sec)

mysql> select '25  Under the age of ' age_cut, count(*) number from user_profile where age < 25 or age is null
    -> union
    -> select '25  Years old and over ' age_cut, count(*) number from user_profile where age >=25;
+-----------------+--------+
| age_cut         | number |
+-----------------+--------+
| 25  Under the age of        |      4 |
| 25  Years old and over      |      3 |
+-----------------+--------+
2 rows in set (0.00 sec)

27、CASE Use of functions

subject ： Divide users into 20 Under the age of ,20-24 year ,25 Three age groups aged and over , View the details of users of different ages , Take out the corresponding data .（ notes ： If age is blank, return other ）

Example ：user_profile The data are as follows .

id	device_id	gender	age	university	gpa	active_days_within_30	question_cnt	answer_cnt
1	2138	male	21	Peking University,	3.4	7	2	12
2	3214	male		Fudan University	4	15	5	25
3	6543	female	20	Peking University,	3.2	12	3	30
4	2315	female	23	Zhejiang University	3.6	5	1	2
5	5432	male	25	Shandong University	3.8	20	15	70
6	2131	male	28	Shandong University	3.3	15	7	13
7	4321	male	26	Fudan University	3.6	9	6	52

The query should return the following results ：

device_id	gender	age_cut
2138	male	20-24 year
3214	male	other
6543	female	20-24 year
2315	female	20-24 year
5432	male	25 Years old and over
2131	male	25 Years old and over
4321	male	25 Years old and over

answer ：

/* select device_id, gender, case when age is null then ' other ' when age >=20 and age <= 24 then '20-24 year ' when age > 24 then '25 Years old and over ' end age_cut from user_profile; */

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

28、 Date function （YEAR） Use

subject ： To calculate the 2021 year 8 Number of user practice questions per day per month , Take out the corresponding data .

Example ：question_practice_detail The data are as follows .

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
……

The query should return the following results ：

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

Table structure and data are as follows ：

/* 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'); */

answer ：

/* 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 Use of functions （1）

subject ： Count the number of contestants of users of each gender , Take out the corresponding results .

Example ：user_submit The data are as follows .

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

The query should return the following results ：

gender	number
male	2
female	3

Table structure and data are as follows ：

/* 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'); */

answer ：

/* 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 Use of functions （2）

subject ：blog_url In the column url The string after the character is the user name of the user's personal blog , Take the user name as a new column , Take out the required data .

Example ：user_submit The data are as follows .

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

The query should return the following results ：

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

answer ：

/* 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 Use of functions （3）

subject ： Count the number of contestants of users of each age , Take out the corresponding results .

Example ：user_submit The data are as follows .

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

The query should return the following results ：

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

answer ：

/* 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)

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Database system principle and Application Tutorial （061）—— MySQL Exercises ： Operation questions 21-31（ 5、 ... and ）

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