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数据库系统原理与应用教程（073）—— MySQL 练习题：操作题 131-140（十七）：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（073）—— MySQL 练习题：操作题 131-140（十七）：综合练习
131、子查询（1）
该题目使用的表和数据如下：
/* drop table if exists film ; drop table if exists category ; drop table if exists film_category ; CREATE TABLE IF NOT EXISTS film ( film_id smallint(5) NOT NULL DEFAULT '0', title varchar(255) NOT NULL, description text, PRIMARY KEY (film_id)); CREATE TABLE category ( category_id tinyint(3) NOT NULL , name varchar(25) NOT NULL, `last_update` timestamp, PRIMARY KEY ( category_id )); CREATE TABLE film_category ( film_id smallint(5) NOT NULL, category_id tinyint(3) NOT NULL, `last_update` timestamp); INSERT INTO film VALUES(1,'ACADEMY DINOSAUR','A Epic Drama of a Feminist And a Mad Scientist who must Battle a Teacher in The Canadian Rockies'); INSERT INTO film VALUES(2,'ACE GOLDFINGER','A Astounding Epistle of a Database Administrator And a Explorer who must Find a Car in Ancient China'); INSERT INTO film VALUES(3,'ADAPTATION HOLES','A Astounding Reflection of a Lumberjack And a Car who must Sink a Lumberjack in A Baloon Factory'); INSERT INTO category VALUES(1,'Action','2006-02-14 20:46:27'); INSERT INTO category VALUES(2,'Animation','2006-02-14 20:46:27'); INSERT INTO category VALUES(3,'Children','2006-02-14 20:46:27'); INSERT INTO category VALUES(4,'Classics','2006-02-14 20:46:27'); INSERT INTO category VALUES(5,'Comedy','2006-02-14 20:46:27'); INSERT INTO category VALUES(6,'Documentary','2006-02-14 20:46:27'); INSERT INTO category VALUES(7,'Drama','2006-02-14 20:46:27'); INSERT INTO category VALUES(8,'Family','2006-02-14 20:46:27'); INSERT INTO category VALUES(9,'Foreign','2006-02-14 20:46:27'); INSERT INTO category VALUES(10,'Games','2006-02-14 20:46:27'); INSERT INTO category VALUES(11,'Horror','2006-02-14 20:46:27'); INSERT INTO film_category VALUES(1,6,'2006-02-14 21:07:09'); INSERT INTO film_category VALUES(2,11,'2006-02-14 21:07:09'); */
电影信息表：film，表中数据如下：
mysql> select * from film;
+---------+------------------+------------------------------------------------------+
| film_id | title | description |
+---------+------------------+------------------------------------------------------+
| 1 | ACADEMY DINOSAUR | A Epic Drama of a Feminist And a Mad Scientist who must
Battle a Teacher in The Canadian Rockies |
| 2 | ACE GOLDFINGER | A Astounding Epistle of a Database Administrator And
a Explorer who must Find a Car in Ancient China |
| 3 | ADAPTATION HOLES | A Astounding Reflection of a Lumberjack And a Car who
must Sink a Lumberjack in A Baloon Factory |
+---------+------------------+------------------------------------------------------+
3 rows in set (0.00 sec)
类别表：category，表中数据如下：
mysql> select * from category;
+-------------+-------------+---------------------+
| category_id | name | last_update |
+-------------+-------------+---------------------+
| 1 | Action | 2006-02-14 20:46:27 |
| 2 | Animation | 2006-02-14 20:46:27 |
| 3 | Children | 2006-02-14 20:46:27 |
| 4 | Classics | 2006-02-14 20:46:27 |
| 5 | Comedy | 2006-02-14 20:46:27 |
| 6 | Documentary | 2006-02-14 20:46:27 |
| 7 | Drama | 2006-02-14 20:46:27 |
| 8 | Family | 2006-02-14 20:46:27 |
| 9 | Foreign | 2006-02-14 20:46:27 |
| 10 | Games | 2006-02-14 20:46:27 |
| 11 | Horror | 2006-02-14 20:46:27 |
+-------------+-------------+---------------------+
11 rows in set (0.00 sec)
电影分类表：film_category，表中数据如下：
mysql> select * from film_category;
+---------+-------------+---------------------+
| film_id | category_id | last_update |
+---------+-------------+---------------------+
| 1 | 6 | 2006-02-14 21:07:09 |
| 2 | 11 | 2006-02-14 21:07:09 |
+---------+-------------+---------------------+
2 rows in set (0.00 sec)
【问题】查询没有分类的电影id 以及电影名称。
解答：
/* select film_id, title from film where film_id not in (select film_id from film_category); */
mysql> select film_id, title from film where film_id not in
-> (select film_id from film_category);
+---------+------------------+
| film_id | title |
+---------+------------------+
| 3 | ADAPTATION HOLES |
+---------+------------------+
1 row in set (0.00 sec)
132、子查询（2）
该题目使用的表和数据如下：
/* drop table if exists film ; drop table if exists category ; drop table if exists film_category ; CREATE TABLE IF NOT EXISTS film ( film_id smallint(5) NOT NULL DEFAULT '0', title varchar(255) NOT NULL, description text, PRIMARY KEY (film_id)); CREATE TABLE category ( category_id tinyint(3) NOT NULL , name varchar(25) NOT NULL, `last_update` timestamp, PRIMARY KEY ( category_id )); CREATE TABLE film_category ( film_id smallint(5) NOT NULL, category_id tinyint(3) NOT NULL, `last_update` timestamp); INSERT INTO film VALUES(1,'ACADEMY DINOSAUR','A Epic Drama of a Feminist And a Mad Scientist who must Battle a Teacher in The Canadian Rockies'); INSERT INTO film VALUES(2,'ACE GOLDFINGER','A Astounding Epistle of a Database Administrator And a Explorer who must Find a Car in Ancient China'); INSERT INTO film VALUES(3,'ADAPTATION HOLES','A Astounding Reflection of a Lumberjack And a Car who must Sink a Lumberjack in A Baloon Factory'); INSERT INTO category VALUES(1,'Action','2006-02-14 20:46:27'); INSERT INTO category VALUES(2,'Animation','2006-02-14 20:46:27'); INSERT INTO category VALUES(3,'Children','2006-02-14 20:46:27'); INSERT INTO category VALUES(4,'Classics','2006-02-14 20:46:27'); INSERT INTO category VALUES(5,'Comedy','2006-02-14 20:46:27'); INSERT INTO category VALUES(6,'Documentary','2006-02-14 20:46:27'); INSERT INTO film_category VALUES(1,1,'2006-02-14 21:07:09'); INSERT INTO film_category VALUES(2,1,'2006-02-14 21:07:09'); INSERT INTO film_category VALUES(3,6,'2006-02-14 21:07:09'); */
电影信息表：film，表中数据如下：
mysql> select * from film;
+---------+------------------+------------------------------------------------------+
| film_id | title | description |
+---------+------------------+------------------------------------------------------+
| 1 | ACADEMY DINOSAUR | A Epic Drama of a Feminist And a Mad Scientist who must
Battle a Teacher in The Canadian Rockies |
| 2 | ACE GOLDFINGER | A Astounding Epistle of a Database Administrator And
a Explorer who must Find a Car in Ancient China |
| 3 | ADAPTATION HOLES | A Astounding Reflection of a Lumberjack And a Car who
must Sink a Lumberjack in A Baloon Factory |
+---------+------------------+------------------------------------------------------+
3 rows in set (0.00 sec)
类别表：category，表中数据如下：
mysql> select * from category;
+-------------+-------------+---------------------+
| category_id | name | last_update |
+-------------+-------------+---------------------+
| 1 | Action | 2006-02-14 20:46:27 |
| 2 | Animation | 2006-02-14 20:46:27 |
| 3 | Children | 2006-02-14 20:46:27 |
| 4 | Classics | 2006-02-14 20:46:27 |
| 5 | Comedy | 2006-02-14 20:46:27 |
| 6 | Documentary | 2006-02-14 20:46:27 |
+-------------+-------------+---------------------+
6 rows in set (0.00 sec)
电影分类表：film_category，表中数据如下：
mysql> select * from film_category;
+---------+-------------+---------------------+
| film_id | category_id | last_update |
+---------+-------------+---------------------+
| 1 | 1 | 2006-02-14 21:07:09 |
| 2 | 1 | 2006-02-14 21:07:09 |
| 3 | 6 | 2006-02-14 21:07:09 |
+---------+-------------+---------------------+
3 rows in set (0.00 sec)
【问题】查询属于 Action 分类的所有电影对应的 title、description。
解答：
/* select title, description from film where film_id in (select film_id from film_category where category_id in (select category_id from category where name = 'Action') ); */
mysql> select title, description from film
-> where film_id in (select film_id from film_category where category_id in
-> (select category_id from category where name = 'Action')
-> );
+------------------+--------------------------------------------------------------+
| title | description |
+------------------+--------------------------------------------------------------+
| ACADEMY DINOSAUR | A Epic Drama of a Feminist And a Mad Scientist who must Battle
a Teacher in The Canadian Rockies |
| ACE GOLDFINGER | A Astounding Epistle of a Database Administrator And a Explorer
who must Find a Car in Ancient China |
+------------------+--------------------------------------------------------------+
2 rows in set (0.00 sec)
133、字符型函数的使用
该题目使用的表和数据如下：
/* drop table if exists `employees` ; CREATE TABLE `employees` ( `emp_no` int(11) NOT NULL, `birth_date` date NOT NULL, `first_name` varchar(14) NOT NULL, `last_name` varchar(16) NOT NULL, `gender` char(1) NOT NULL, `hire_date` date NOT NULL, PRIMARY KEY (`emp_no`)); INSERT INTO employees VALUES(10001,'1953-09-02','Georgi','Facello','M','1986-06-26'); INSERT INTO employees VALUES(10002,'1964-06-02','Bezalel','Simmel','F','1985-11-21'); INSERT INTO employees VALUES(10003,'1959-12-03','Parto','Bamford','M','1986-08-28'); INSERT INTO employees VALUES(10004,'1954-05-01','Chirstian','Koblick','M','1986-12-01'); INSERT INTO employees VALUES(10005,'1955-01-21','Kyoichi','Maliniak','M','1989-09-12'); INSERT INTO employees VALUES(10006,'1953-04-20','Anneke','Preusig','F','1989-06-02'); INSERT INTO employees VALUES(10007,'1957-05-23','Tzvetan','Zielinski','F','1989-02-10'); INSERT INTO employees VALUES(10008,'1958-02-19','Saniya','Kalloufi','M','1994-09-15'); INSERT INTO employees VALUES(10009,'1952-04-19','Sumant','Peac','F','1985-02-18'); INSERT INTO employees VALUES(10010,'1963-06-01','Duangkaew','Piveteau','F','1989-08-24'); INSERT INTO employees VALUES(10011,'1953-11-07','Mary','Sluis','F','1990-01-22'); */
员工表：employees，表中数据如下：
mysql> select * from employees;
+--------+------------+------------+-----------+--------+------------+
| emp_no | birth_date | first_name | last_name | gender | hire_date |
+--------+------------+------------+-----------+--------+------------+
| 10001 | 1953-09-02 | Georgi | Facello | M | 1986-06-26 |
| 10002 | 1964-06-02 | Bezalel | Simmel | F | 1985-11-21 |
| 10003 | 1959-12-03 | Parto | Bamford | M | 1986-08-28 |
| 10004 | 1954-05-01 | Chirstian | Koblick | M | 1986-12-01 |
| 10005 | 1955-01-21 | Kyoichi | Maliniak | M | 1989-09-12 |
| 10006 | 1953-04-20 | Anneke | Preusig | F | 1989-06-02 |
| 10007 | 1957-05-23 | Tzvetan | Zielinski | F | 1989-02-10 |
| 10008 | 1958-02-19 | Saniya | Kalloufi | M | 1994-09-15 |
| 10009 | 1952-04-19 | Sumant | Peac | F | 1985-02-18 |
| 10010 | 1963-06-01 | Duangkaew | Piveteau | F | 1989-08-24 |
| 10011 | 1953-11-07 | Mary | Sluis | F | 1990-01-22 |
+--------+------------+------------+-----------+--------+------------+
11 rows in set (0.00 sec)
【问题】将 employees 表的所有员工的 last_name 和 first_name 拼接起来作为 Name，中间以一个空格区分。查询结果如下：
| Name |
|---|
| Facello Georgi |
| Simmel Bezalel |
| Bamford Parto |
| Koblick Chirstian |
| Maliniak Kyoichi |
| Preusig Anneke |
| Zielinski Tzvetan |
| Kalloufi Saniya |
| Peac Sumant |
| Piveteau Duangkaew |
| Sluis Mary |
解答：
mysql> select concat(last_name,' ',first_name) name from employees;
+--------------------+
| name |
+--------------------+
| Facello Georgi |
| Simmel Bezalel |
| Bamford Parto |
| Koblick Chirstian |
| Maliniak Kyoichi |
| Preusig Anneke |
| Zielinski Tzvetan |
| Kalloufi Saniya |
| Peac Sumant |
| Piveteau Duangkaew |
| Sluis Mary |
+--------------------+
11 rows in set (0.00 sec)
134、查询时强制使用索引
该题目使用的表和数据如下：
/* drop table if exists salaries; CREATE TABLE `salaries` ( `emp_no` int(11) NOT NULL, `salary` int(11) NOT NULL, `from_date` date NOT NULL, `to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`from_date`)); create index idx_emp_no on salaries(emp_no); INSERT INTO salaries VALUES(10005,78228,'1989-09-12','1990-09-12'); INSERT INTO salaries VALUES(10005,94692,'2001-09-09','9999-01-01'); */
【问题】查询 emp_no 为10005 的记录，强制使用索引 idx_emp_no。
解答：
mysql> explain SELECT * FROM salaries FORCE INDEX (idx_emp_no)
WHERE emp_no='10005'\G
*************************** 1. row ***************************
id: 1
select_type: SIMPLE
table: salaries
partitions: NULL
type: ref
possible_keys: idx_emp_no
key: idx_emp_no
key_len: 4
ref: const
rows: 2
filtered: 100.00
Extra: NULL
1 row in set, 1 warning (0.00 sec)
135、删除重复记录
该题目使用的表和数据如下：
/* drop table if exists titles_test; CREATE TABLE titles_test ( id int(11) not null primary key, emp_no int(11) NOT NULL, title varchar(50) NOT NULL, from_date date NOT NULL, to_date date DEFAULT NULL); insert into titles_test values ('1', '10001', 'Senior Engineer', '1986-06-26', '9999-01-01'), ('2', '10002', 'Staff', '1996-08-03', '9999-01-01'), ('3', '10003', 'Senior Engineer', '1995-12-03', '9999-01-01'), ('4', '10004', 'Senior Engineer', '1995-12-03', '9999-01-01'), ('5', '10001', 'Senior Engineer', '1986-06-26', '9999-01-01'), ('6', '10002', 'Staff', '1996-08-03', '9999-01-01'), ('7', '10003', 'Senior Engineer', '1995-12-03', '9999-01-01'); */
mysql> select * from titles_test;
+----+--------+-----------------+------------+------------+
| id | emp_no | title | from_date | to_date |
+----+--------+-----------------+------------+------------+
| 1 | 10001 | Senior Engineer | 1986-06-26 | 9999-01-01 |
| 2 | 10002 | Staff | 1996-08-03 | 9999-01-01 |
| 3 | 10003 | Senior Engineer | 1995-12-03 | 9999-01-01 |
| 4 | 10004 | Senior Engineer | 1995-12-03 | 9999-01-01 |
| 5 | 10001 | Senior Engineer | 1986-06-26 | 9999-01-01 |
| 6 | 10002 | Staff | 1996-08-03 | 9999-01-01 |
| 7 | 10003 | Senior Engineer | 1995-12-03 | 9999-01-01 |
+----+--------+-----------------+------------+------------+
7 rows in set (0.00 sec)
【问题】删除 emp_no 重复的记录，只保留最小的 id 对应的记录。查询结果如下：
+----+--------+-----------------+------------+------------+
| id | emp_no | title | from_date | to_date |
+----+--------+-----------------+------------+------------+
| 1 | 10001 | Senior Engineer | 1986-06-26 | 9999-01-01 |
| 2 | 10002 | Staff | 1996-08-03 | 9999-01-01 |
| 3 | 10003 | Senior Engineer | 1995-12-03 | 9999-01-01 |
| 4 | 10004 | Senior Engineer | 1995-12-03 | 9999-01-01 |
+----+--------+-----------------+------------+------------+
解答：
/* DELETE FROM titles_test WHERE id NOT IN (SELECT MIN(id) FROM titles_test GROUP BY emp_no); */
mysql> DELETE FROM titles_test WHERE id NOT IN
-> (SELECT MIN(id) FROM titles_test GROUP BY emp_no);
ERROR 1093 (HY000): You can't specify target table 'titles_test' for update in FROM clause'
-- 出错，mysql 的 delete from 语句的子查询中不能出现要删除的表名。
mysql> create table titles_test_bak select * FROM titles_test WHERE id IN
-> (SELECT MIN(id) FROM titles_test GROUP BY emp_no);
Query OK, 4 rows affected (0.09 sec)
Records: 4 Duplicates: 0 Warnings: 0
mysql> truncate table titles_test;
Query OK, 0 rows affected (0.04 sec)
mysql> insert into titles_test select * from titles_test_bak;
Query OK, 4 rows affected (0.00 sec)
Records: 4 Duplicates: 0 Warnings: 0
mysql>
mysql> select * from titles_test;
+----+--------+-----------------+------------+------------+
| id | emp_no | title | from_date | to_date |
+----+--------+-----------------+------------+------------+
| 1 | 10001 | Senior Engineer | 1986-06-26 | 9999-01-01 |
| 2 | 10002 | Staff | 1996-08-03 | 9999-01-01 |
| 3 | 10003 | Senior Engineer | 1995-12-03 | 9999-01-01 |
| 4 | 10004 | Senior Engineer | 1995-12-03 | 9999-01-01 |
+----+--------+-----------------+------------+------------+
4 rows in set (0.00 sec)
136、使用 replace into 命令更新数据
该题目使用的表和数据如下：
/* drop table if exists titles_test; CREATE TABLE titles_test ( id int(11) not null primary key, emp_no int(11) NOT NULL, title varchar(50) NOT NULL, from_date date NOT NULL, to_date date DEFAULT NULL); insert into titles_test values ('1', '10001', 'Senior Engineer', '1986-06-26', '9999-01-01'), ('2', '10002', 'Staff', '1996-08-03', '9999-01-01'), ('3', '10003', 'Senior Engineer', '1995-12-03', '9999-01-01'), ('4', '10004', 'Senior Engineer', '1995-12-03', '9999-01-01'), ('5', '10001', 'Senior Engineer', '1986-06-26', '9999-01-01'), ('6', '10002', 'Staff', '1996-08-03', '9999-01-01'), ('7', '10003', 'Senior Engineer', '1995-12-03', '9999-01-01'); */
mysql> select * from titles_test;
+----+--------+-----------------+------------+------------+
| id | emp_no | title | from_date | to_date |
+----+--------+-----------------+------------+------------+
| 1 | 10001 | Senior Engineer | 1986-06-26 | 9999-01-01 |
| 2 | 10002 | Staff | 1996-08-03 | 9999-01-01 |
| 3 | 10003 | Senior Engineer | 1995-12-03 | 9999-01-01 |
| 4 | 10004 | Senior Engineer | 1995-12-03 | 9999-01-01 |
| 5 | 10001 | Senior Engineer | 1986-06-26 | 9999-01-01 |
| 6 | 10002 | Staff | 1996-08-03 | 9999-01-01 |
| 7 | 10003 | Senior Engineer | 1995-12-03 | 9999-01-01 |
+----+--------+-----------------+------------+------------+
7 rows in set (0.00 sec)
【问题】将 id = 5 以及 emp_no = 10001 的行数据替换成 id = 5 以及 emp_no = 10005，其他数据保持不变。必须使用replace 实现，不能使用 update 。
解答：
/* replace into titles_test select 5,10005,title,from_date,to_date from titles_test where id=5 and emp_no=10001; */
mysql> replace into titles_test
-> select 5,10005,title,from_date,to_date
-> from titles_test where id=5 and emp_no=10001;
Query OK, 2 rows affected (0.02 sec)
Records: 1 Duplicates: 1 Warnings: 0
mysql> select * from titles_test;
+----+--------+-----------------+------------+------------+
| id | emp_no | title | from_date | to_date |
+----+--------+-----------------+------------+------------+
| 1 | 10001 | Senior Engineer | 1986-06-26 | 9999-01-01 |
| 2 | 10002 | Staff | 1996-08-03 | 9999-01-01 |
| 3 | 10003 | Senior Engineer | 1995-12-03 | 9999-01-01 |
| 4 | 10004 | Senior Engineer | 1995-12-03 | 9999-01-01 |
| 5 | 10005 | Senior Engineer | 1986-06-26 | 9999-01-01 |
| 6 | 10002 | Staff | 1996-08-03 | 9999-01-01 |
| 7 | 10003 | Senior Engineer | 1995-12-03 | 9999-01-01 |
+----+--------+-----------------+------------+------------+
7 rows in set (0.00 sec)
137、字符型函数的使用
该题目使用的表和数据如下：
/* drop table if exists strings; CREATE TABLE strings( id int(5) NOT NULL PRIMARY KEY, string varchar(200) NOT NULL ); insert into strings values (1, '10,A,B,郑州,河南'), (2, 'A,B,C,D,新乡，河南'), (3, '中国河南新乡A,11,B,C,D,E'); */
mysql> select * from strings;
+----+--------------------------------+
| id | string |
+----+--------------------------------+
| 1 | 10,A,B,郑州,河南 |
| 2 | A,B,C,D,新乡，河南 |
| 3 | 中国河南新乡A,11,B,C,D,E |
+----+--------------------------------+
3 rows in set (0.00 sec)
【问题】统计每个字符串中逗号出现的次数 cnt。查询结果如下：
| id | cnt |
|---|---|
| 1 | 4 |
| 2 | 4 |
| 3 | 5 |
解答：
/* select id, char_length(string)-char_length(replace(string,',','')) cnt from strings; */
mysql> select id, char_length(string)-char_length(replace(string,',','')) cnt
-> from strings;
+----+------+
| id | cnt |
+----+------+
| 1 | 4 |
| 2 | 4 |
| 3 | 5 |
+----+------+
3 rows in set (0.05 sec)
138、子查询
该题目使用的表和数据如下：
/* drop table if exists `salaries` ; CREATE TABLE `salaries` ( `emp_no` int(11) NOT NULL, `salary` float(11,3) NOT NULL, `from_date` date NOT NULL, `to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`from_date`)); INSERT INTO salaries VALUES(10001,85097,'2001-06-22','2002-06-22'); INSERT INTO salaries VALUES(10001,88958,'2002-06-22','9999-01-01'); INSERT INTO salaries VALUES(10002,72527,'2001-08-02','9999-01-01'); INSERT INTO salaries VALUES(10003,43699,'2000-12-01','2001-12-01'); INSERT INTO salaries VALUES(10003,43311,'2001-12-01','9999-01-01'); INSERT INTO salaries VALUES(10004,70698,'2000-11-27','2001-11-27'); INSERT INTO salaries VALUES(10004,74057,'2001-11-27','9999-01-01'); */
mysql> select * from salaries;
+--------+-----------+------------+------------+
| emp_no | salary | from_date | to_date |
+--------+-----------+------------+------------+
| 10001 | 85097.000 | 2001-06-22 | 2002-06-22 |
| 10001 | 88958.000 | 2002-06-22 | 9999-01-01 |
| 10002 | 72527.000 | 2001-08-02 | 9999-01-01 |
| 10003 | 43699.000 | 2000-12-01 | 2001-12-01 |
| 10003 | 43311.000 | 2001-12-01 | 9999-01-01 |
| 10004 | 70698.000 | 2000-11-27 | 2001-11-27 |
| 10004 | 74057.000 | 2001-11-27 | 9999-01-01 |
+--------+-----------+------------+------------+
7 rows in set (0.00 sec)
【问题】查询排除在职（to_date = ‘9999-01-01’）员工的最大、最小 salary 之后，其他的在职员工的平均工资 avg_salary。查询结果如下：
| avg_salary |
|---|
| 73292 |
解答：
/* select avg(salary) avg_salary from salaries where to_date = '9999-01-01' and salary not in (select max(salary) from salaries where to_date = '9999-01-01') and salary not in (select min(salary) from salaries where to_date = '9999-01-01'); */
mysql> select avg(salary) avg_salary
-> from salaries where to_date = '9999-01-01' and salary not in
-> (select max(salary) from salaries where to_date = '9999-01-01')
-> and salary not in
-> (select min(salary) from salaries where to_date = '9999-01-01');
+---------------+
| avg_salary |
+---------------+
| 73292.0000000 |
+---------------+
1 row in set (0.02 sec)
139、exits 子查询
该题目使用的表和数据如下：
/* drop table if exists employees; drop table if exists dept_emp; CREATE TABLE `employees` ( `emp_no` int(11) NOT NULL, `birth_date` date NOT NULL, `first_name` varchar(14) NOT NULL, `last_name` varchar(16) NOT NULL, `gender` char(1) NOT NULL, `hire_date` date NOT NULL, PRIMARY KEY (`emp_no`)); CREATE TABLE `dept_emp` ( `emp_no` int(11) NOT NULL, `dept_no` char(4) NOT NULL, `from_date` date NOT NULL, `to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`dept_no`)); INSERT INTO employees VALUES(10001,'1953-09-02','Georgi','Facello','M','1986-06-26'); INSERT INTO employees VALUES(10002,'1964-06-02','Bezalel','Simmel','F','1985-11-21'); INSERT INTO employees VALUES(10003,'1959-12-03','Parto','Bamford','M','1986-08-28'); INSERT INTO employees VALUES(10004,'1954-05-01','Chirstian','Koblick','M','1986-12-01'); INSERT INTO employees VALUES(10005,'1955-01-21','Kyoichi','Maliniak','M','1989-09-12'); INSERT INTO employees VALUES(10006,'1953-04-20','Anneke','Preusig','F','1989-06-02'); INSERT INTO employees VALUES(10007,'1957-05-23','Tzvetan','Zielinski','F','1989-02-10'); INSERT INTO employees VALUES(10008,'1958-02-19','Saniya','Kalloufi','M','1994-09-15'); INSERT INTO employees VALUES(10009,'1952-04-19','Sumant','Peac','F','1985-02-18'); INSERT INTO employees VALUES(10010,'1963-06-01','Duangkaew','Piveteau','F','1989-08-24'); INSERT INTO employees VALUES(10011,'1953-11-07','Mary','Sluis','F','1990-01-22'); INSERT INTO dept_emp VALUES(10001,'d001','1986-06-26','9999-01-01'); INSERT INTO dept_emp VALUES(10002,'d001','1996-08-03','9999-01-01'); INSERT INTO dept_emp VALUES(10003,'d004','1995-12-03','9999-01-01'); INSERT INTO dept_emp VALUES(10004,'d004','1986-12-01','9999-01-01'); INSERT INTO dept_emp VALUES(10005,'d003','1989-09-12','9999-01-01'); INSERT INTO dept_emp VALUES(10006,'d002','1990-08-05','9999-01-01'); INSERT INTO dept_emp VALUES(10007,'d005','1989-02-10','9999-01-01'); INSERT INTO dept_emp VALUES(10008,'d005','1998-03-11','2000-07-31'); INSERT INTO dept_emp VALUES(10009,'d006','1985-02-18','9999-01-01'); INSERT INTO dept_emp VALUES(10010,'d005','1996-11-24','2000-06-26'); INSERT INTO dept_emp VALUES(10010,'d006','2000-06-26','9999-01-01'); */
dept_emp 表中数据如下：
mysql> select * from dept_emp;
+--------+---------+------------+------------+
| emp_no | dept_no | from_date | to_date |
+--------+---------+------------+------------+
| 10001 | d001 | 1986-06-26 | 9999-01-01 |
| 10002 | d001 | 1996-08-03 | 9999-01-01 |
| 10003 | d004 | 1995-12-03 | 9999-01-01 |
| 10004 | d004 | 1986-12-01 | 9999-01-01 |
| 10005 | d003 | 1989-09-12 | 9999-01-01 |
| 10006 | d002 | 1990-08-05 | 9999-01-01 |
| 10007 | d005 | 1989-02-10 | 9999-01-01 |
| 10008 | d005 | 1998-03-11 | 2000-07-31 |
| 10009 | d006 | 1985-02-18 | 9999-01-01 |
| 10010 | d005 | 1996-11-24 | 2000-06-26 |
| 10010 | d006 | 2000-06-26 | 9999-01-01 |
+--------+---------+------------+------------+
11 rows in set (0.00 sec)
employees 表中数据如下：
mysql> select * from employees;
+--------+------------+------------+-----------+--------+------------+
| emp_no | birth_date | first_name | last_name | gender | hire_date |
+--------+------------+------------+-----------+--------+------------+
| 10001 | 1953-09-02 | Georgi | Facello | M | 1986-06-26 |
| 10002 | 1964-06-02 | Bezalel | Simmel | F | 1985-11-21 |
| 10003 | 1959-12-03 | Parto | Bamford | M | 1986-08-28 |
| 10004 | 1954-05-01 | Chirstian | Koblick | M | 1986-12-01 |
| 10005 | 1955-01-21 | Kyoichi | Maliniak | M | 1989-09-12 |
| 10006 | 1953-04-20 | Anneke | Preusig | F | 1989-06-02 |
| 10007 | 1957-05-23 | Tzvetan | Zielinski | F | 1989-02-10 |
| 10008 | 1958-02-19 | Saniya | Kalloufi | M | 1994-09-15 |
| 10009 | 1952-04-19 | Sumant | Peac | F | 1985-02-18 |
| 10010 | 1963-06-01 | Duangkaew | Piveteau | F | 1989-08-24 |
| 10011 | 1953-11-07 | Mary | Sluis | F | 1990-01-22 |
+--------+------------+------------+-----------+--------+------------+
11 rows in set (0.00 sec)
【问题】使用含有关键字 exists 查找未分配具体部门的员工的所有信息。查询结果如下：
| emp_no | birth_date | first_name | last_name | gender | hire_date |
|---|---|---|---|---|---|
| 10011 | 1953-11-07 | Mary | Sluis | F | 1990-01-22 |
解答：
/* select * from employees where not exists (select * from dept_emp where emp_no = employees.emp_no); */
mysql> select * from employees where not exists
-> (select * from dept_emp where emp_no = employees.emp_no);
+--------+------------+------------+-----------+--------+------------+
| emp_no | birth_date | first_name | last_name | gender | hire_date |
+--------+------------+------------+-----------+--------+------------+
| 10011 | 1953-11-07 | Mary | Sluis | F | 1990-01-22 |
+--------+------------+------------+-----------+--------+------------+
1 row in set (0.01 sec)
140、case 函数的用法
该题目使用的表和数据如下：
/* drop table if exists `employees` ; drop table if exists emp_bonus; drop table if exists `salaries` ; CREATE TABLE `employees` ( `emp_no` int(11) NOT NULL, `birth_date` date NOT NULL, `first_name` varchar(14) NOT NULL, `last_name` varchar(16) NOT NULL, `gender` char(1) NOT NULL, `hire_date` date NOT NULL, PRIMARY KEY (`emp_no`)); create table emp_bonus( emp_no int not null, recevied datetime not null, btype smallint not null); CREATE TABLE `salaries` ( `emp_no` int(11) NOT NULL, `salary` int(11) NOT NULL, `from_date` date NOT NULL, `to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`from_date`)); insert into emp_bonus values (10001, '2010-01-01',1), (10002, '2010-10-01',2); INSERT INTO employees VALUES(10001,'1953-09-02','Georgi','Facello','M','1986-06-26'); INSERT INTO employees VALUES(10002,'1964-06-02','Bezalel','Simmel','F','1985-11-21'); INSERT INTO salaries VALUES(10001,60117,'1986-06-26','1987-06-26'); INSERT INTO salaries VALUES(10001,62102,'1987-06-26','1988-06-25'); INSERT INTO salaries VALUES(10001,66074,'1988-06-25','1989-06-25'); INSERT INTO salaries VALUES(10001,66596,'1989-06-25','1990-06-25'); INSERT INTO salaries VALUES(10001,66961,'1990-06-25','1991-06-25'); INSERT INTO salaries VALUES(10001,71046,'1991-06-25','1992-06-24'); INSERT INTO salaries VALUES(10001,74333,'1992-06-24','1993-06-24'); INSERT INTO salaries VALUES(10001,75286,'1993-06-24','1994-06-24'); INSERT INTO salaries VALUES(10001,75994,'1994-06-24','1995-06-24'); INSERT INTO salaries VALUES(10001,76884,'1995-06-24','1996-06-23'); INSERT INTO salaries VALUES(10001,80013,'1996-06-23','1997-06-23'); INSERT INTO salaries VALUES(10001,81025,'1997-06-23','1998-06-23'); INSERT INTO salaries VALUES(10001,81097,'1998-06-23','1999-06-23'); INSERT INTO salaries VALUES(10001,84917,'1999-06-23','2000-06-22'); INSERT INTO salaries VALUES(10001,85112,'2000-06-22','2001-06-22'); INSERT INTO salaries VALUES(10001,85097,'2001-06-22','2002-06-22'); INSERT INTO salaries VALUES(10001,88958,'2002-06-22','9999-01-01'); INSERT INTO salaries VALUES(10002,72527,'1996-08-03','1997-08-03'); INSERT INTO salaries VALUES(10002,72527,'1997-08-03','1998-08-03'); INSERT INTO salaries VALUES(10002,72527,'1998-08-03','1999-08-03'); INSERT INTO salaries VALUES(10002,72527,'1999-08-03','2000-08-02'); INSERT INTO salaries VALUES(10002,72527,'2000-08-02','2001-08-02'); INSERT INTO salaries VALUES(10002,72527,'2001-08-02','9999-01-01'); */
员工表：employees，表中数据如下：
mysql> select * from employees;
+--------+------------+------------+-----------+--------+------------+
| emp_no | birth_date | first_name | last_name | gender | hire_date |
+--------+------------+------------+-----------+--------+------------+
| 10001 | 1953-09-02 | Georgi | Facello | M | 1986-06-26 |
| 10002 | 1964-06-02 | Bezalel | Simmel | F | 1985-11-21 |
+--------+------------+------------+-----------+--------+------------+
2 rows in set (0.00 sec)
员工奖金表：emp_bonus，表中数据如下：
mysql> select * from emp_bonus;
+--------+---------------------+-------+
| emp_no | recevied | btype |
+--------+---------------------+-------+
| 10001 | 2010-01-01 00:00:00 | 1 |
| 10002 | 2010-10-01 00:00:00 | 2 |
+--------+---------------------+-------+
2 rows in set (0.00 sec)
薪水表：salaries，表中数据如下：
mysql> select * from salaries;
+--------+--------+------------+------------+
| emp_no | salary | from_date | to_date |
+--------+--------+------------+------------+
| 10001 | 60117 | 1986-06-26 | 1987-06-26 |
| 10001 | 62102 | 1987-06-26 | 1988-06-25 |
| 10001 | 66074 | 1988-06-25 | 1989-06-25 |
| 10001 | 66596 | 1989-06-25 | 1990-06-25 |
| 10001 | 66961 | 1990-06-25 | 1991-06-25 |
| 10001 | 71046 | 1991-06-25 | 1992-06-24 |
| 10001 | 74333 | 1992-06-24 | 1993-06-24 |
| 10001 | 75286 | 1993-06-24 | 1994-06-24 |
| 10001 | 75994 | 1994-06-24 | 1995-06-24 |
| 10001 | 76884 | 1995-06-24 | 1996-06-23 |
| 10001 | 80013 | 1996-06-23 | 1997-06-23 |
| 10001 | 81025 | 1997-06-23 | 1998-06-23 |
| 10001 | 81097 | 1998-06-23 | 1999-06-23 |
| 10001 | 84917 | 1999-06-23 | 2000-06-22 |
| 10001 | 85112 | 2000-06-22 | 2001-06-22 |
| 10001 | 85097 | 2001-06-22 | 2002-06-22 |
| 10001 | 88958 | 2002-06-22 | 9999-01-01 |
| 10002 | 72527 | 1996-08-03 | 1997-08-03 |
| 10002 | 72527 | 1997-08-03 | 1998-08-03 |
| 10002 | 72527 | 1998-08-03 | 1999-08-03 |
| 10002 | 72527 | 1999-08-03 | 2000-08-02 |
| 10002 | 72527 | 2000-08-02 | 2001-08-02 |
| 10002 | 72527 | 2001-08-02 | 9999-01-01 |
+--------+--------+------------+------------+
23 rows in set (0.00 sec)
【问题】emp_bonus 表中 btype 取值为 1，表示奖金为薪水 salary 的 10%，btype 为 2 表示奖金为薪水的 20%，其他类型奖金均为薪水的 30%。to_date = ‘9999-01-01’ 表示当前薪水。
查询 emp_no，first_name，last_name，奖金类型 btype、对应的当前薪水情况（salary）以及奖金金额（bonus）。bonus 结果保留一位小数，输出结果按 emp_no 升序排序。查询结果如下：
| emp_no | first_name | last_name | btype | salary | bonus |
|---|---|---|---|---|---|
| 10001 | Georgi | Facello | 1 | 88958 | 8895.8000 |
| 10002 | Bezalel | Simmel | 2 | 72527 | 14505.4000 |
解答：
/* select e.emp_no, e.first_name, e.last_name, eb.btype, s.salary, round((case btype when 1 then salary * 0.1 when 2 then salary * 0.2 else salary * 0.3 end),1) bonus from employees e join emp_bonus eb on e.emp_no = eb.emp_no join salaries s on e.emp_no = s.emp_no where s.to_date = '9999-01-01' order by e.emp_no; */
mysql> select e.emp_no, e.first_name, e.last_name, eb.btype, s.salary,
-> (case btype when 1 then salary * 0.1 when 2 then salary * 0.2 else salary * 0.3 end) bonus
-> from employees e join emp_bonus eb on e.emp_no = eb.emp_no
-> join salaries s on e.emp_no = s.emp_no
-> where s.to_date = '9999-01-01';
+--------+------------+-----------+-------+--------+---------+
| emp_no | first_name | last_name | btype | salary | bonus |
+--------+------------+-----------+-------+--------+---------+
| 10001 | Georgi | Facello | 1 | 88958 | 8895.8 |
| 10002 | Bezalel | Simmel | 2 | 72527 | 14505.4 |
+--------+------------+-----------+-------+--------+---------+
2 rows in set (0.00 sec)
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