牛客网数据库SQL实战(16-20)

2022-12-11 07:14:38

16、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);

select titles.title,avg(salaries.salary) as avg

from titles

join salaries

on titles.emp_no = salaries.emp_no

and titles.to_date = '9999-01-01'

and salaries.to_date = '9999-01-01'

group by titles.title;

select titles.title,avg(salaries.salary) as avg

from titles

inner join salaries

on titles.emp_no = salaries.emp_no

and titles.to_date = '9999-01-01'

and salaries.to_date = '9999-01-01'

group by titles.title;

讨论：https://www.nowcoder.com/questionTerminal/c8652e9e5a354b879e2a244200f1eaae

17、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`));

select emp_no,salary

from salaries

where to_date = '9999-01-01'

order by salary desc

limit 1,1;

讨论：https://www.nowcoder.com/questionTerminal/8d2c290cc4e24403b98ca82ce45d04db

18、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`));

SELECT e.emp_no, MAX(s.salary) AS salary, e.last_name, e.first_name

FROM employees AS e

INNER JOIN salaries AS s

ON e.emp_no = s.emp_no

WHERE s.to_date = '9999-01-01'

AND s.salary NOT IN

(SELECT MAX(salary) FROM salaries WHERE to_date = '9999-01-01');

讨论：https://www.nowcoder.com/questionTerminal/c1472daba75d4635b7f8540b837cc719

19、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`));

select employees.last_name,employees.first_name,departments.dept_name

from employees

left join dept_emp

on employees.emp_no = dept_emp.emp_no

left join departments

on dept_emp.dept_no = departments.dept_no;

讨论：https://www.nowcoder.com/questionTerminal/5a7975fabe1146329cee4f670c27ad55

20、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`));

select (max(salary) - min(salary)) as growth

from salaries

where emp_no = 10001;

讨论：https://www.nowcoder.com/questionTerminal/c727647886004942a89848e2b5130dc2

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