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);
- 查找入职员工时间排名倒数第三的员工所有信息
CREATE TABLEemployees
(
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);
- 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 TABLEdept_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 TABLEsalaries
(
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;
- 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 TABLEdept_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 TABLEemployees
(
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;
- 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 TABLEemployees
(
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 TABLEsalaries
(
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 ;
- 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 TABLEsalaries
(
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;
- 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 TABLEsalaries
(
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
- 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 TABLEdept_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 TABLEsalaries
(
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;
- 获取所有非manager的员工emp_no
CREATE TABLEdept_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 TABLEemployees
(
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;