数据库SQL实战-获取所有部门中当前员工薪水最高的相关信息(mysql)

其他 2021-03-04 05:51:47 阅读次数: 0

1. 获取所有员工当前的manager

1.1 题目描述

获取所有员工当前的(dept_manager.to_date=‘9999-01-01’)manager，如果员工是manager的话不显示(也就是如果当前的manager是自己的话结果不显示)。输出结果第一列给出当前员工的emp_no,第二列给出其manager对应的emp_no。
CREATE TABLE dept_emp (
emp_no int(11) NOT NULL, – ‘所有的员工编号’
dept_no char(4) NOT NULL, – ‘部门编号’
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, – ‘部门编号’
emp_no int(11) NOT NULL, – ‘经理编号’
from_date date NOT NULL,
to_date date NOT NULL,
PRIMARY KEY (emp_no,dept_no));
如插入：
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’);

INSERT INTO dept_manager VALUES(‘d001’,10002,‘1996-08-03’,‘9999-01-01’);
INSERT INTO dept_manager VALUES(‘d002’,10006,‘1990-08-05’,‘9999-01-01’);
INSERT INTO dept_manager VALUES(‘d003’,10005,‘1989-09-12’,‘9999-01-01’);
INSERT INTO dept_manager VALUES(‘d004’,10004,‘1986-12-01’,‘9999-01-01’);
INSERT INTO dept_manager VALUES(‘d005’,10010,‘1996-11-24’,‘2000-06-26’);
INSERT INTO dept_manager VALUES(‘d006’,10010,‘2000-06-26’,‘9999-01-01’);

输入描述：

无

输出描述：

在这里插入图片描述

1.2 语句实现

select e.emp_no, m.emp_no as manager_no
from dept_emp e left join dept_manager m
on e.dept_no = m.dept_no
where e.emp_no != m.emp_no and m.to_date = '9999-01-01'

2. 获取所有部门中当前员工薪水最高的相关信息

2.1 题目描述

获取所有部门中当前(dept_emp.to_date = ‘9999-01-01’)员工当前(salaries.to_date=‘9999-01-01’)薪水最高的相关信息，给出dept_no, emp_no以及其对应的salary，按照部门编号升序排列。
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));
如插入：
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,‘d001’,‘1996-08-03’,‘1997-08-03’);

INSERT INTO salaries VALUES(10001,90000,‘1986-06-26’,‘1987-06-26’);
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,‘2000-08-02’,‘2001-08-02’);
INSERT INTO salaries VALUES(10002,72527,‘2001-08-02’,‘9999-01-01’);
INSERT INTO salaries VALUES(10003,90000,‘1996-08-03’,‘1997-08-03’);
则输出
在这里插入图片描述

2.2 语句实现

select e.dept_no, e.emp_no, s.salary
from dept_emp e inner join salaries s
on e.emp_no = s.emp_no and e.to_date = '9999-01-01' and s.to_date = '9999-01-01'
where s.salary in (select max(s1.salary)
    from dept_emp e1 inner join salaries s1
    on e1.emp_no = s1.emp_no and e1.to_date = '9999-01-01'
    and s1.to_date = '9999-01-01' and e1.dept_no = e.dept_no
)
order by e.dept_no

3. 从titles表获取按照title进行分组，注意对于重复的emp_no进行忽略

3.1 题目描述

从titles表获取按照title进行分组，每组个数大于等于2，给出title以及对应的数目t。
CREATE TABLE IF NOT EXISTS “titles” (
emp_no int(11) NOT NULL,
title varchar(50) NOT NULL,
from_date date NOT NULL,
to_date date DEFAULT NULL);
如插入：
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’);
INSERT INTO titles VALUES(10004,‘Engineer’,‘1986-12-01’,‘1995-12-01’);
INSERT INTO titles VALUES(10004,‘Senior Engineer’,‘1995-12-01’,‘9999-01-01’);
INSERT INTO titles VALUES(10005,‘Senior Staff’,‘1996-09-12’,‘9999-01-01’);
INSERT INTO titles VALUES(10005,‘Staff’,‘1989-09-12’,‘1996-09-12’);
INSERT INTO titles VALUES(10006,‘Senior Engineer’,‘1990-08-05’,‘9999-01-01’);
INSERT INTO titles VALUES(10007,‘Senior Staff’,‘1996-02-11’,‘9999-01-01’);
INSERT INTO titles VALUES(10007,‘Staff’,‘1989-02-10’,‘1996-02-11’);
INSERT INTO titles VALUES(10008,‘Assistant Engineer’,‘1998-03-11’,‘2000-07-31’);
INSERT INTO titles VALUES(10009,‘Assistant Engineer’,‘1985-02-18’,‘1990-02-18’);
INSERT INTO titles VALUES(10009,‘Engineer’,‘1990-02-18’,‘1995-02-18’);
INSERT INTO titles VALUES(10009,‘Senior Engineer’,‘1995-02-18’,‘9999-01-01’);
INSERT INTO titles VALUES(10010,‘Engineer’,‘1996-11-24’,‘9999-01-01’);
INSERT INTO titles VALUES(10010,‘Engineer’,‘1996-11-24’,‘9999-01-01’);

输入描述：

无

输出描述：

在这里插入图片描述

3.2 语句实现

select t.title, count(t.emp_no) as t
from titles t
group by t.title
having t >= 2

如果对于重复的emp_no进行忽略（即emp_no重复的title不计算，title对应的数目t不增加），则语句实现如下：

select t.title, count(distinct t.emp_no) as t
from titles t
group by t.title
having t >= 2

4. 查找employees表所有emp_no为奇数

4.1 题目描述

查找employees表所有emp_no为奇数，且last_name不为Mary(注意大小写)的员工信息，并按照hire_date逆序排列
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));
如插入：
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’);

输入描述：

无

输出描述：

在这里插入图片描述

4.2 语句实现

select *
from employees
where emp_no % 2 = 1 and last_name != 'Mary'
order by hire_date desc

5. 统计出当前各个title类型对应的员工当前薪水对应的平均工资

5.1 题目描述

统计出当前(titles.to_date=‘9999-01-01’)各个title类型对应的员工当前(salaries.to_date=‘9999-01-01’)薪水对应的平均工资。结果给出title以及平均工资avg。
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 IF NOT EXISTS “titles” (
emp_no int(11) NOT NULL,
title varchar(50) NOT NULL,
from_date date NOT NULL,
to_date date DEFAULT NULL);

如插入：
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(10004,70698,‘1986-12-01’,‘1995-12-01’);
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,‘Engineer’,‘1986-12-01’,‘1995-12-01’);
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’);

输出：

在这里插入图片描述

5.2 语句实现

select t.title, avg(s.salary)
from titles t join salaries s on t.emp_no = s.emp_no
and t.to_date = '9999-01-01' 
and s.to_date = '9999-01-01'
group by title

6. 查找当前薪水排名第二多的员工编号emp_no

6.1 题目描述

获取当前（to_date=‘9999-01-01’）薪水第二多的员工的emp_no以及其对应的薪水salary
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));

进阶：查找当前薪水(to_date=‘9999-01-01’)排名第二多的员工编号emp_no、薪水salary、last_name以及first_name，你可以不使用order by完成吗？
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));

6.2 语句实现

/*
select s.emp_no, s.salary
from salaries s
where s.to_date = '9999-01-01'
order by s.salary desc
limit 1,1
*/

#进阶
select e.emp_no, s.salary, e.last_name, e.first_name
from employees e inner join salaries s 
on e.emp_no = s.emp_no
where s.to_date = '9999-01-01'
and s.salary = (select max(salary)
               from salaries
               where to_date = '9999-01-01'
               and salary < (select max(salary)
                             from salaries 
                             where to_date = '9999-01-01'
                            )
               )

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