牛客网数据库SQL实战(21-25)

2022-10-22 21:55:14

21、查找所有员工自入职以来的薪水涨幅情况，给出员工编号emp_no以及其对应的薪水涨幅growth，并按照growth进行升序
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 a.emp_no, (b.salary - c.salary) as growth

from

    employees as a

    inner join salaries as b

    on a.emp_no = b.emp_no and b.to_date = '9999-01-01'

    inner join salaries as c

    on a.emp_no = c.emp_no and a.hire_date = c.from_date

order by growth asc;

讨论：https://www.nowcoder.com/questionTerminal/fc7344ece7294b9e98401826b94c6ea5

22、统计各个部门对应员工涨幅的次数总和，给出部门编码dept_no、部门名称dept_name以及次数sum
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`));

select departments.dept_no,departments.dept_name,count(*) sum

from departments,dept_emp,salaries

where departments.dept_no = dept_emp.dept_no

and dept_emp.emp_no = salaries.emp_no

group by departments.dept_no;

讨论：https://www.nowcoder.com/questionTerminal/6a62b6c0a7324350a6d9959fa7c21db3

23、对所有员工的当前(to_date='9999-01-01')薪水按照salary进行按照1-N的排名，相同salary并列且按照emp_no升序排列
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 s1.emp_no,s1.salary,count(distinct s2.salary)

from salaries s1,salaries s2

where s1.to_date='9999-01-01' and s2.to_date='9999-01-01' and s1.salary<= s2.salary/*给定s1.salary求有多少个s2.salary大于它*/

group by s1.emp_no

order by s1.salary desc,s1.emp_no asc；

讨论：https://www.nowcoder.com/questionTerminal/b9068bfe5df74276bd015b9729eec4bf

24、获取所有非manager员工当前的薪水情况，给出dept_no、emp_no以及salary ，当前表示to_date='9999-01-01'
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`));

SELECT de.dept_no, s.emp_no, s.salary

FROM dept_emp AS de

INNER JOIN salaries AS s

ON s.emp_no = de.emp_no

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

WHERE de.emp_no NOT IN

(SELECT emp_no FROM dept_manager WHERE to_date = '9999-01-01');

1、先用INNER JOIN连接employees和salaries，找出当前所有员工的工资情况

2、再用INNER JOIN连接dept_emp表，找到所有员工所在的部门

3、最后用限制条件de.emp_no NOT IN (SELECT emp_no FROM dept_manager WHERE to_date = '9999-01-01')选出当前所有非manager员工，再依次输出dept_no、emp_no、salary

讨论：https://www.nowcoder.com/questionTerminal/8fe212a6c71b42de9c15c56ce354bebe

25、获取员工其当前的薪水比其manager当前薪水还高的相关信息，当前表示to_date='9999-01-01',
结果第一列给出员工的emp_no，
第二列给出其manager的manager_no，
第三列给出该员工当前的薪水emp_salary,
第四列给该员工对应的manager当前的薪水manager_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 `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`));

SELECT sem.emp_no AS emp_no, sdm.emp_no AS manager_no, sem.salary AS emp_salary, sdm.salary AS manager_salary

FROM

(SELECT s.salary, s.emp_no, de.dept_no

 FROM salaries s

 INNER JOIN dept_emp de

 ON s.emp_no = de.emp_no

 AND s.to_date = '9999-01-01' ) AS sem,

(SELECT s.salary, s.emp_no, dm.dept_no

 FROM salaries s

 INNER JOIN dept_manager dm

 ON s.emp_no = dm.emp_no

 AND s.to_date = '9999-01-01' ) AS sdm

WHERE sem.dept_no = sdm.dept_no

AND sem.salary > sdm.salary;

1、先用INNER JOIN连接salaries和demp_emp，建立当前所有员工的工资记录sem

2、再用INNER JOIN连接salaries和demp_manager，建立当前所有员工的工资记录sdm

3、最后用限制条件sem.dept_no = sdm.dept_no AND sem.salary > sdm.salary找出同一部门中工资比经理高的员工，并根据题意依次输出emp_no、manager_no、emp_salary、manager_salary

讨论：https://www.nowcoder.com/questionTerminal/f858d74a030e48da8e0f69e21be63bef

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