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数据库系统原理与应用教程（072）—— MySQL 练习题：操作题 121-130（十六）：综合练习
2022-08-01 19:13:00 【睿思达DBA_WGX】
数据库系统原理与应用教程（072）—— MySQL 练习题：操作题 121-130（十六）：综合练习
121、分组查询
该题目使用的表和数据如下：
/* drop table if exists `salaries` ; drop table if exists titles; 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 TABLE titles ( `emp_no` int(11) NOT NULL, `title` varchar(50) NOT NULL, `from_date` date NOT NULL, `to_date` date DEFAULT NULL); INSERT INTO salaries VALUES(10001,88958,'1986-06-26','9999-01-01'); INSERT INTO salaries VALUES(10003,43311,'2001-12-01','9999-01-01'); INSERT INTO salaries VALUES(10004,74057,'1995-12-01','9999-01-01'); INSERT INTO salaries VALUES(10006,43311,'2001-08-02','9999-01-01'); INSERT INTO salaries VALUES(10007,88070,'2002-02-07','9999-01-01'); INSERT INTO titles VALUES(10001,'Senior Engineer','1986-06-26','9999-01-01'); INSERT INTO titles VALUES(10003,'Senior Engineer','2001-12-01','9999-01-01'); INSERT INTO titles VALUES(10004,'Senior Engineer','1995-12-01','9999-01-01'); INSERT INTO titles VALUES(10006,'Senior Engineer','2001-08-02','9999-01-01'); INSERT INTO titles VALUES(10007,'Senior Staff','1996-02-11','9999-01-01'); */
员工职称表：titles，表中数据如下：
mysql> select * from titles;
+--------+-----------------+------------+------------+
| emp_no | title | from_date | to_date |
+--------+-----------------+------------+------------+
| 10001 | Senior Engineer | 1986-06-26 | 9999-01-01 |
| 10003 | Senior Engineer | 2001-12-01 | 9999-01-01 |
| 10004 | Senior Engineer | 1995-12-01 | 9999-01-01 |
| 10006 | Senior Engineer | 2001-08-02 | 9999-01-01 |
| 10007 | Senior Staff | 1996-02-11 | 9999-01-01 |
+--------+-----------------+------------+------------+
5 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 | 88958 | 1986-06-26 | 9999-01-01 |
| 10003 | 43311 | 2001-12-01 | 9999-01-01 |
| 10004 | 74057 | 1995-12-01 | 9999-01-01 |
| 10006 | 43311 | 2001-08-02 | 9999-01-01 |
| 10007 | 88070 | 2002-02-07 | 9999-01-01 |
+--------+--------+------------+------------+
5 rows in set (0.00 sec)
【问题】查询各个 title 类型对应员工的平均工资。显示 title 以及平均工资，并且以平均工资升序排序，查询结果如下：
| title | avg(s.salary) |
|---|---|
| Senior Engineer | 62409.2500 |
| Senior Staff | 88070.0000 |
解答：
/* select t.title, avg(s.salary) from titles t join salaries s on t.emp_no = s.emp_no group by t.title order by 2; */
mysql> select t.title, avg(s.salary)
-> from titles t join salaries s on t.emp_no = s.emp_no
-> group by t.title
-> order by 2;
+-----------------+---------------+
| title | avg(s.salary) |
+-----------------+---------------+
| Senior Engineer | 62409.2500 |
| Senior Staff | 88070.0000 |
+-----------------+---------------+
2 rows in set (0.00 sec)
122、子查询（1）
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`)); 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(10003,43311,'2001-12-01','9999-01-01'); */
薪水表：salaries，表中数据如下：
mysql> select * from salaries;
+--------+--------+------------+------------+
| emp_no | salary | from_date | to_date |
+--------+--------+------------+------------+
| 10001 | 88958 | 2002-06-22 | 9999-01-01 |
| 10002 | 72527 | 2001-08-02 | 9999-01-01 |
| 10003 | 43311 | 2001-12-01 | 9999-01-01 |
+--------+--------+------------+------------+
3 rows in set (0.00 sec)
【问题】查询薪水第二多的员工（如有重复则全部取出）的 emp_no 以及其对应的薪水 salary，并按 emp_no 升序排序。查询结果如下：
| emp_no | salary |
|---|---|
| 10002 | 72527 |
解答：
/* select emp_no, salary from salaries where salary in (select salary from (select salary from salaries order by salary desc limit 1,1) a) order by emp_no; */
mysql> select emp_no, salary from salaries where salary in
-> (select salary from (select salary from salaries order by salary desc limit 1,1) a);
+--------+--------+
| emp_no | salary |
+--------+--------+
| 10002 | 72527 |
+--------+--------+
1 row in set (0.00 sec)
123、子查询（2）
该题目使用的表和数据如下：
/* drop table if exists `employees` ; 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 `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 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 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(10003,43311,'2001-12-01','9999-01-01'); INSERT INTO salaries VALUES(10004,74057,'2001-11-27','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 |
| 10003 | 1959-12-03 | Parto | Bamford | M | 1986-08-28 |
| 10004 | 1954-05-01 | Chirstian | Koblick | M | 1986-12-01 |
+--------+------------+------------+-----------+--------+------------+
4 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 | 88958 | 2002-06-22 | 9999-01-01 |
| 10002 | 72527 | 2001-08-02 | 9999-01-01 |
| 10003 | 43311 | 2001-12-01 | 9999-01-01 |
| 10004 | 74057 | 2001-11-27 | 9999-01-01 |
+--------+--------+------------+------------+
4 rows in set (0.00 sec)
【问题】查询薪水排名第二多的员工编号 emp_no、薪水 salary、last_name 以及 first_name，查询结果如下：
| emp_no | salary | last_name | first_name |
|---|---|---|---|
| 10004 | 74057 | Koblick | Chirstian |
/* select e.emp_no, s.salary, e.last_name, e.first_name from employees e join salaries s on e.emp_no = s.emp_no where s.salary in (select max(salary) from salaries where salary < (select max(salary) from salaries)); */
mysql> select e.emp_no, s.salary, e.last_name, e.first_name
-> from employees e join salaries s on e.emp_no = s.emp_no
-> where s.salary in
-> (select max(salary) from salaries where salary < (select max(salary) from salaries));
+--------+--------+-----------+------------+
| emp_no | salary | last_name | first_name |
+--------+--------+-----------+------------+
| 10004 | 74057 | Koblick | Chirstian |
+--------+--------+-----------+------------+
1 row in set (0.04 sec)
124、连接查询（1）
该题目使用的表和数据如下：
/* drop table if exists `departments` ; drop table if exists `dept_emp` ; drop table if exists `employees` ; CREATE TABLE `departments` ( `dept_no` char(4) NOT NULL, `dept_name` varchar(40) NOT NULL, PRIMARY KEY (`dept_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`)); 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 departments VALUES('d001','Marketing'); INSERT INTO departments VALUES('d002','Finance'); INSERT INTO departments VALUES('d003','Human Resources'); 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,'d002','1990-08-05','9999-01-01'); 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'); */
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 |
+--------+------------+------------+-----------+--------+------------+
4 rows in set (0.00 sec)
部门表：departments，表中数据如下：
mysql> select * from departments;
+---------+-----------------+
| dept_no | dept_name |
+---------+-----------------+
| d001 | Marketing |
| d002 | Finance |
| d003 | Human Resources |
+---------+-----------------+
3 rows in set (0.00 sec)
部门员工关系表：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 | d002 | 1990-08-05 | 9999-01-01 |
+--------+---------+------------+------------+
3 rows in set (0.00 sec)
【问题】查询所有员工的 last_name 和 first_name 以及对应的 dept_name，包括暂时没有分配部门的员工，查询结果如下：
| last_name | first_name | dept_name |
|---|---|---|
| Facello | Georgi | Marketing |
| Simmel | Bezalel | Marketing |
| Bamford | Parto | Finance |
| Koblick | Chirstian | NULL |
解答：
/* select e.last_name, e.first_name, d.dept_name from dept_emp de join departments d on de.dept_no = d.dept_no right join employees e on e.emp_no = de.emp_no; */
mysql> select e.last_name, e.first_name, d.dept_name
-> from dept_emp de join departments d on de.dept_no = d.dept_no
-> right join employees e on e.emp_no = de.emp_no;
+-----------+------------+-----------+
| last_name | first_name | dept_name |
+-----------+------------+-----------+
| Facello | Georgi | Marketing |
| Simmel | Bezalel | Marketing |
| Bamford | Parto | Finance |
| Koblick | Chirstian | NULL |
+-----------+------------+-----------+
4 rows in set (0.00 sec)
125、连接查询（2）
该题目使用的表和数据如下：
/* drop table if exists `employees` ; 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 `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 employees VALUES(10001,'1953-09-02','Georgi','Facello','M','2001-06-22'); INSERT INTO employees VALUES(10002,'1964-06-02','Bezalel','Simmel','F','1999-08-03'); 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,'1999-08-03','2000-08-02'); INSERT INTO salaries VALUES(10002,72527,'2000-08-02','2001-08-02'); */
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 | 2001-06-22 |
| 10002 | 1964-06-02 | Bezalel | Simmel | F | 1999-08-03 |
+--------+------------+------------+-----------+--------+------------+
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 | 85097 | 2001-06-22 | 2002-06-22 |
| 10001 | 88958 | 2002-06-22 | 9999-01-01 |
| 10002 | 72527 | 1999-08-03 | 2000-08-02 |
| 10002 | 72527 | 2000-08-02 | 2001-08-02 |
+--------+--------+------------+------------+
4 rows in set (0.00 sec)
【问题】查询在职员工自入职以来的薪水涨幅情况，给出在职员工编号 emp_no 以及其对应的薪水涨幅 growth，并按照growth 进行升序排列，查询结果如下：
| emp_no | growth |
|---|---|
| 10001 | 3861 |
解答：
/* select e.emp_no, a.salary - s.salary growth from employees e join salaries s on e.emp_no = s.emp_no and e.hire_date = s.from_date join (select emp_no, salary from salaries where to_date = '9999-01-01') a on s.emp_no = a.emp_no order by growth; */
mysql> select e.emp_no, a.salary - s.salary growth
-> from employees e join salaries s on e.emp_no = s.emp_no and e.hire_date = s.from_date
-> join (select emp_no, salary from salaries where to_date = '9999-01-01') a
-> on s.emp_no = a.emp_no order by growth;
+--------+--------+
| emp_no | growth |
+--------+--------+
| 10001 | 3861 |
+--------+--------+
1 row in set (0.00 sec)
126、连接查询（3）
该题目使用的表和数据如下：
/* drop table if exists `departments` ; drop table if exists `dept_emp` ; drop table if exists `salaries` ; CREATE TABLE `departments` ( `dept_no` char(4) NOT NULL, `dept_name` varchar(40) NOT NULL, PRIMARY KEY (`dept_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`)); 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 departments VALUES('d001','Marketing'); INSERT INTO departments VALUES('d002','Finance'); INSERT INTO dept_emp VALUES(10001,'d001','2001-06-22','9999-01-01'); INSERT INTO dept_emp VALUES(10002,'d001','1996-08-03','9999-01-01'); INSERT INTO dept_emp VALUES(10003,'d002','1996-08-03','9999-01-01'); 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','9999-01-01'); INSERT INTO salaries VALUES(10003,32323,'1996-08-03','9999-01-01'); */
部门表：departments，表中数据如下：
mysql> select * from departments;
+---------+-----------+
| dept_no | dept_name |
+---------+-----------+
| d001 | Marketing |
| d002 | Finance |
+---------+-----------+
2 rows in set (0.00 sec)
部门员工关系表：dept_emp，表中数据如下：
mysql> select * from dept_emp;
+--------+---------+------------+------------+
| emp_no | dept_no | from_date | to_date |
+--------+---------+------------+------------+
| 10001 | d001 | 2001-06-22 | 9999-01-01 |
| 10002 | d001 | 1996-08-03 | 9999-01-01 |
| 10003 | d002 | 1996-08-03 | 9999-01-01 |
+--------+---------+------------+------------+
3 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 | 85097 | 2001-06-22 | 2002-06-22 |
| 10001 | 88958 | 2002-06-22 | 9999-01-01 |
| 10002 | 72527 | 1996-08-03 | 9999-01-01 |
| 10003 | 32323 | 1996-08-03 | 9999-01-01 |
+--------+--------+------------+------------+
4 rows in set (0.00 sec)
【问题】统计各个部门的工资记录数，给出部门编码（dept_no），部门名称（dept_name）以及部门在 salaries 表里面有多少条记录（sum），按照 dept_no 升序排序，查询结果如下：
| dept_no | dept_name | sum |
|---|---|---|
| d001 | Marketing | 3 |
| d002 | Finance | 1 |
解答：
/* select d.dept_no, d.dept_name, count(*) sum from salaries s join dept_emp de on s.emp_no = de.emp_no join departments d on de.dept_no = d.dept_no group by d.dept_no, d.dept_name order by d.dept_no; */
mysql> select d.dept_no, d.dept_name, count(*) sum
-> from salaries s join dept_emp de on s.emp_no = de.emp_no
-> join departments d on de.dept_no = d.dept_no
-> group by d.dept_no, d.dept_name
-> order by d.dept_no;
+---------+-----------+-----+
| dept_no | dept_name | sum |
+---------+-----------+-----+
| d001 | Marketing | 3 |
| d002 | Finance | 1 |
+---------+-----------+-----+
2 rows in set (0.00 sec)
127、查询数据排名
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`)); 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(10003,43311,'2001-12-01','9999-01-01'); INSERT INTO salaries VALUES(10004,72527,'2001-12-01','9999-01-01'); */
薪水表：salaries，表中数据如下：
mysql> select * from salaries;
+--------+--------+------------+------------+
| emp_no | salary | from_date | to_date |
+--------+--------+------------+------------+
| 10001 | 88958 | 2002-06-22 | 9999-01-01 |
| 10002 | 72527 | 2001-08-02 | 9999-01-01 |
| 10003 | 43311 | 2001-12-01 | 9999-01-01 |
| 10004 | 72527 | 2001-12-01 | 9999-01-01 |
+--------+--------+------------+------------+
4 rows in set (0.00 sec)
【问题】对所有员工的薪水按照 salary 降序进行排名，如果 salary 相同，再按照 emp_no 升序排列，查询结果如下：
| emp_no | salary | t_rank |
|---|---|---|
| 10001 | 88958 | 1 |
| 10002 | 72527 | 2 |
| 10004 | 72527 | 2 |
| 10003 | 43311 | 3 |
解答：
/* select emp_no, salary, @rank := @rank + (CASE WHEN @salary = salary THEN 0 WHEN @salary := salary THEN 1 END) t_rank from salaries,(select @rank := 0) a,(select @salary := null) b order by salary desc, emp_no; */
mysql> select emp_no, salary,
-> @rank := @rank + (CASE WHEN @salary = salary THEN 0 WHEN @salary := salary THEN 1 END) t_rank
-> from salaries,(select @rank := 0) a,(select @salary := null) b
-> order by salary desc, emp_no;
+--------+--------+--------+
| emp_no | salary | t_rank |
+--------+--------+--------+
| 10001 | 88958 | 1 |
| 10002 | 72527 | 2 |
| 10004 | 72527 | 2 |
| 10003 | 43311 | 3 |
+--------+--------+--------+
4 rows in set (0.00 sec)
128、连接查询
该题目使用的表和数据如下：
/* drop table if exists `dept_emp` ; drop table if exists `dept_manager` ; drop table if exists `employees` ; drop table if exists `salaries` ; 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`)); CREATE TABLE `dept_manager` ( `dept_no` char(4) NOT NULL, `emp_no` int(11) NOT NULL, `from_date` date NOT NULL, `to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`dept_no`)); 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 `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 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_manager VALUES('d001',10002,'1996-08-03','9999-01-01'); 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','1996-08-03'); INSERT INTO salaries VALUES(10001,88958,'1986-06-26','9999-01-01'); INSERT INTO salaries VALUES(10002,72527,'1996-08-03','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 | 1996-08-03 |
+--------+------------+------------+-----------+--------+------------+
2 rows in set (0.00 sec)
部门员工关系表：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 |
+--------+---------+------------+------------+
2 rows in set (0.00 sec)
部门经理表：dept_manager，表中数据如下：
mysql> select * from dept_manager;
+---------+--------+------------+------------+
| dept_no | emp_no | from_date | to_date |
+---------+--------+------------+------------+
| d001 | 10002 | 1996-08-03 | 9999-01-01 |
+---------+--------+------------+------------+
1 row 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 | 88958 | 1986-06-26 | 9999-01-01 |
| 10002 | 72527 | 1996-08-03 | 9999-01-01 |
+--------+--------+------------+------------+
2 rows in set (0.00 sec)
【问题】查询所有非 manager 员工的薪水情况，显示：dept_no，emp_no 以及 salary，查询结果如下：
| dept_no | emp_no | salary |
|---|---|---|
| d001 | 10001 | 88958 |
/* select de.dept_no, e.emp_no, s.salary from dept_emp de join employees e on de.emp_no = e.emp_no join salaries s on e.emp_no = s.emp_no where e.emp_no not in (select emp_no from dept_manager); */
mysql> select de.dept_no, e.emp_no, s.salary
-> from dept_emp de join employees e on de.emp_no = e.emp_no
-> join salaries s on e.emp_no = s.emp_no
-> where e.emp_no not in (select emp_no from dept_manager);
+---------+--------+--------+
| dept_no | emp_no | salary |
+---------+--------+--------+
| d001 | 10001 | 88958 |
+---------+--------+--------+
1 row in set (0.00 sec)
129、子查询与连接查询
该题目使用的表和数据如下：
/* drop table if exists `dept_emp` ; drop table if exists `dept_manager` ; drop table if exists `salaries` ; 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`)); CREATE TABLE `dept_manager` ( `dept_no` char(4) NOT NULL, `emp_no` int(11) NOT NULL, `from_date` date NOT NULL, `to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`dept_no`)); 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 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_manager VALUES('d001',10002,'1996-08-03','9999-01-01'); INSERT INTO salaries VALUES(10001,88958,'2002-06-22','9999-01-01'); INSERT INTO salaries VALUES(10002,72527,'1996-08-03','9999-01-01'); */
部门关系表：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 |
+--------+---------+------------+------------+
2 rows in set (0.00 sec)
部门经理表：dept_manager，表中数据如下：
mysql> select * from dept_manager;
+---------+--------+------------+------------+
| dept_no | emp_no | from_date | to_date |
+---------+--------+------------+------------+
| d001 | 10002 | 1996-08-03 | 9999-01-01 |
+---------+--------+------------+------------+
1 row 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 | 88958 | 2002-06-22 | 9999-01-01 |
| 10002 | 72527 | 1996-08-03 | 9999-01-01 |
+--------+--------+------------+------------+
2 rows in set (0.00 sec)
【问题】查询员工的的薪水比其 manager 当前薪水还高的相关信息，显示员工的 emp_no，对应的 manager 的manager_no，该员工当前的薪水 emp_salary，该员工对应的 manager 当前的薪水 manager_salary。查询结果如下：
| emp_no | manager_no | emp_salary | manager_salary |
|---|---|---|---|
| 10001 | 10002 | 88958 | 72527 |
解答：
/* select emp.emp_no, manager.emp_no, emp.emp_salary, manager.manager_salary from (select de.emp_no, de.dept_no, s.salary emp_salary from dept_emp de join salaries s on de.emp_no = s.emp_no) emp join (select dm.dept_no, dm.emp_no, s.salary manager_salary from dept_manager dm join salaries s on dm.emp_no = s.emp_no) manager on emp.dept_no = manager.dept_no where emp.emp_salary > manager.manager_salary; */
mysql> select emp.emp_no, manager.emp_no, emp.emp_salary, manager.manager_salary
-> from (select de.emp_no, de.dept_no, s.salary emp_salary
-> from dept_emp de join salaries s on de.emp_no = s.emp_no) emp join
-> (select dm.dept_no, dm.emp_no, s.salary manager_salary
-> from dept_manager dm join salaries s on dm.emp_no = s.emp_no) manager
-> on emp.dept_no = manager.dept_no
-> where emp.emp_salary > manager.manager_salary;
+--------+--------+------------+----------------+
| emp_no | emp_no | emp_salary | manager_salary |
+--------+--------+------------+----------------+
| 10001 | 10002 | 88958 | 72527 |
+--------+--------+------------+----------------+
1 row in set (0.00 sec)
130、连接查询与分组
该题目使用的表和数据如下：
/* drop table if exists `departments` ; drop table if exists `dept_emp` ; drop table if exists titles ; CREATE TABLE `departments` ( `dept_no` char(4) NOT NULL, `dept_name` varchar(40) NOT NULL, PRIMARY KEY (`dept_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`)); CREATE TABLE titles ( `emp_no` int(11) NOT NULL, `title` varchar(50) NOT NULL, `from_date` date NOT NULL, `to_date` date DEFAULT NULL); INSERT INTO departments VALUES('d001','Marketing'); INSERT INTO departments VALUES('d002','Finance'); 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,'d002','1995-12-03','9999-01-01'); INSERT INTO titles VALUES(10001,'Senior Engineer','1986-06-26','9999-01-01'); INSERT INTO titles VALUES(10002,'Staff','1996-08-03','9999-01-01'); INSERT INTO titles VALUES(10003,'Senior Engineer','1995-12-03','9999-01-01'); */
部门表：departments，表中数据如下：
mysql> select * from departments;
+---------+-----------+
| dept_no | dept_name |
+---------+-----------+
| d001 | Marketing |
| d002 | Finance |
+---------+-----------+
2 rows in set (0.00 sec)
部门员工关系表：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 | d002 | 1995-12-03 | 9999-01-01 |
+--------+---------+------------+------------+
3 rows in set (0.00 sec)
职称表：titles，表中数据如下：
mysql> select * from titles;
+--------+-----------------+------------+------------+
| emp_no | title | from_date | to_date |
+--------+-----------------+------------+------------+
| 10001 | Senior Engineer | 1986-06-26 | 9999-01-01 |
| 10002 | Staff | 1996-08-03 | 9999-01-01 |
| 10003 | Senior Engineer | 1995-12-03 | 9999-01-01 |
+--------+-----------------+------------+------------+
3 rows in set (0.00 sec)
【问题】汇总各个部门当前员工的 title 类型的分配数目，查询结果包括：部门编号（dept_no），dept_name，部门下所有的员工的 title 以及该类型 title 对应的数目 count，结果按照 dept_no 升序排序，dept_no 一样的再按 title 升序排序。查询结果如下：
| dept_no | dept_name | title | count |
|---|---|---|---|
| d001 | Marketing | Senior Engineer | 1 |
| d001 | Marketing | Staff | 1 |
| d002 | Finance | Senior Engineer | 1 |
解答：
/* select d.dept_no, d.dept_name, t.title, count(*) count from dept_emp de join departments d on de.dept_no = d.dept_no join titles t on de.emp_no = t.emp_no group by d.dept_no, d.dept_name, t.title order by d.dept_no, title; */
mysql> select d.dept_no, d.dept_name, t.title, count(*) count
-> from dept_emp de join departments d on de.dept_no = d.dept_no
-> join titles t on de.emp_no = t.emp_no
-> group by d.dept_no, d.dept_name, t.title
-> order by d.dept_no, title;
+---------+-----------+-----------------+-------+
| dept_no | dept_name | title | count |
+---------+-----------+-----------------+-------+
| d001 | Marketing | Senior Engineer | 1 |
| d001 | Marketing | Staff | 1 |
| d002 | Finance | Senior Engineer | 1 |
+---------+-----------+-----------------+-------+
3 rows in set (0.00 sec)
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