牛客网SQL题目答案

---

title: 牛客网SQL题目答案
date: 2018-4-20 21:18:40
categories:
- MySQL
tags:

- MySQL

摘要： NEWCODER SQL

牛客SQL一直用的就是那几张表
1. 查找最晚入职员工的所有信息
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));

sql:

select * from employees where hire_date = (select max(hire_date) from employees);

1. 查找入职员工时间排名倒数第三的员工所有信息
   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 * from employees where hire_date = (select hire_date from employees order by hire_date desc limit 2,1);

1. 查找各个部门当前(to_date=’9999-01-01’)领导当前薪水详情以及其对应部门编号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 s.*,d.dept_no from salaries s ,dept_manager d where s.emp_no = d.emp_no and d.to_date='9999-01-01' and s.to_date='9999-01-01';

4. 查找所有已经分配部门的员工的last_name和first_name
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 e.last_name, e.first_name,d.dept_no from dept_emp d,employees e where d.emp_no = e.emp_no;

1. 查找所有员工的last_name和first_name以及对应部门编号dept_no，也包括展示没有分配具体部门的员工
   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 e.last_name, e.first_name ,d.dept_no from employees e left join dept_emp d on e.emp_no = d.emp_no;

1. 查找所有员工入职时候的薪水情况，给出emp_no以及salary， 并按照emp_no进行逆序
   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 e.emp_no, s.salary from employees e, salaries s
where e.emp_no = s.emp_no and e.hire_date = s.from_date
order by e.emp_no DESC ;

1. 查找薪水涨幅超过15次的员工号emp_no以及其对应的涨幅次数t
   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 salaries.emp_no, count(salaries.emp_no) t from salaries
group by salaries.emp_no having t > 15;

1. 找出所有员工当前(to_date=’9999-01-01’)具体的薪水salary情况，对于相同的薪水只显示一次,并按照逆序显示
   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 distinct salaries.salary from salaries where salaries.to_date = '9999-01-01'
order by salaries.salary DESC

1. 获取所有部门当前manager的当前薪水情况，给出dept_no, emp_no以及salary，当前表示to_date=’9999-01-01’
   CREATE TABLE dept_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 dept_manager.dept_no, dept_manager.emp_no, salaries.salary from salaries , dept_manager
where salaries.to_date='9999-01-01' and dept_manager.to_date='9999-01-01' and dept_manager.emp_no = salaries.emp_no;

1. 获取所有非manager的员工emp_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));

明显可以使用很多方法来写SQl
* 使用NOT IN选出在employees但不在dept_manager中的emp_no记录
SELECT emp_no FROM employees
WHERE emp_no NOT IN (SELECT emp_no FROM dept_manager)
* 先使用LEFT JOIN连接两张表，再从此表中选出dept_no值为NULL对应的emp_no记录

select e.emp_no from employees e left join dept_manager d
on e.emp_no = d.emp_no where d.dept_no is null;