牛客网—数据库SQL实战（2）
11. 获取所有员工当前的manager
获取所有员工当前的manager，如果当前的manager是自己的话结果不显示，当前表示to_date='9999-01-01'。
结果第一列给出当前员工的emp_no,第二列给出其manager对应的manager_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 `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`));
输入描述:
无输出描述:
| emp_no | manager_no |
|---|---|
| 10001 | 10002 |
| 10003 | 10004 |
| 10009 | 10010 |
select d1.emp_no, d2.emp_no as manager_no
from dept_emp as d1, dept_manager as d2
where d1.dept_no = d2.dept_no
and d1.emp_no <> d2.emp_no
and d1.to_date = '9999-01-01'
and d2.to_date = '9999-01-01'12. 获取所有部门中当前员工薪水最高的相关信息
题目描述
获取所有部门中当前员工薪水最高的相关信息，给出dept_no, emp_no以及其对应的salary
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`));
输入描述:
无输出描述:
| dept_no | emp_no | salary |
|---|---|---|
| d001 | 10001 | 88958 |
| d002 | 10006 | 43311 |
| d003 | 10005 | 94692 |
| d004 | 10004 | 74057 |
| d005 | 10007 | 88070 |
| d006 | 10009 | 95409 |
select d.dept_no, d.emp_no, max(s.salary) as salary
from dept_emp as d, salaries as s
where d.emp_no = s.emp_no
and d.to_date = '9999-01-01'
and s.to_date = '9999-01-01'
group by d.dept_no13. 从titles表获取按照title进行分组
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);
输入描述:
无输出描述:
| title | t |
|---|---|
| Assistant Engineer | 2 |
| Engineer | 4 |
| 省略 | 省略 |
| Staff | 3 |
select title, count(title) as t
from titles
group by title
having t >= 214. 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
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。
注意对于重复的emp_no进行忽略。
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);
输入描述:
无
输出描述:
| title | t |
|---|---|
| Assistant Engineer | 2 |
| Engineer | 3 |
| 省略 | 省略 |
| Staff | 3 |
select title, count(distinct emp_no) as t
from titles
group by title
having t >= 215. 查找employees表
题目描述
查找employees表所有emp_no为奇数，且last_name不为Mary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`));
输入描述:
无输出描述:
| emp_no | birth_date | first_name | last_name | gender | hire_date |
|---|---|---|---|---|---|
| 10011 | 1953-11-07 | Mary | Sluis | F | 1990-01-22 |
| 10005 | 1955-01-21 | Kyoichi | Maliniak | M | 1989-09-12 |
| 10007 | 1957-05-23 | Tzvetan | Zielinski | F | 1989-02-10 |
| 10003 | 1959-12-03 | Parto | Bamford | M | 1986-08-28 |
| 10001 | 1953-09-02 | Georgi | Facello | M | 1986-06-26 |
| 10009 | 1952-04-19 | Sumant | Peac | F | 1985-02-18 |
select emp_no, birth_date, first_name, last_name, gender, hire_date
from employees
where emp_no % 2 = 1
and last_name <> 'Mary'
order by hire_date desc16. 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
题目描述
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。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);
输入描述:
无输出描述:
| title | avg |
|---|---|
| Engineer | 94409.0 |
| Senior Engineer | 69009.2 |
| Senior Staff | 91381.0 |
| Staff | 72527.0 |
select t.title, avg(s.salary) as avg
from titles as t, salaries as s
where t.emp_no = s.emp_no
and t.to_date = '9999-01-01'
and s.to_date = '9999-01-01'
group by t.title17. 获取当前薪水第二多的员工的emp_no以及其对应的薪水salary
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`));
输入描述:
无输出描述:
| emp_no | salary |
|---|---|
| 10009 | 94409 |
select emp_no, salary
from salaries
order by salary desc limit 1,118. 查找当前薪水排名第二多的员工编号emp_no以及其对应的薪水salary，不准使用order_by
题目描述
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，不准使用order by
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`));
输入描述:
无输出描述:
| emp_no | salary | last_name | first_name |
|---|---|---|---|
| 10009 | 94409 | Peac | Sumant |
select e.emp_no, s.salary, e.last_name, e.first_name
from employees as e, salaries as s
where e.emp_no = s.emp_no
and 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"))19. 查找所有员工的last_name和first_name以及对应的dept_name
题目描述
查找所有员工的last_name和first_name以及对应的dept_name，也包括暂时没有分配部门的员工
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`));
输入描述:
无输出描述:
| last_name | first_name | dept_name |
|---|---|---|
| Facello | Georgi | Marketing |
| 省略 | 省略 | 省略 |
| Sluis | Mary | NULL |
select e.last_name, e.first_name, d2.dept_name
from employees as e left join dept_emp as d1
on e.emp_no = d1.emp_no left join departments as d2
on d1.dept_no = d2.dept_no20. 查找员工编号emp_no为10001其自入职以来的薪水salary涨幅值growth
题目描述
查找员工编号emp_no为10001其自入职以来的薪水salary涨幅值growth
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`));
输入描述:
无输出描述:
| growth |
|---|
| 28841 |
select (
(select salary from salaries where emp_no = 10001 order by to_date desc limit 1) -
(select salary from salaries where emp_no = 10001 order by to_date asc limit 1)
) as growth