SQL语句训练6-10

题目地址
https://www.nowcoder.com/activity/oj

6.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));

这题存在坑点，一个人的工资记录可能不止一个，存在涨工资的可能
这里利用from_date和hire_date来确定入职时的工资

select t1.emp_no,salary 
from employees t1
inner join salaries t2
on t1.emp_no=t2.emp_no and t1.hire_date=t2.from_date
order by t1.emp_no desc;

7.题目描述
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));

按工号聚合后计算工资记录数，大于15的算一条

select emp_no, count(salary) as t 
from salaries 
group by emp_no 
having t > 15;

8.题目描述
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 salary 
from salaries 
where to_date = '9999-01-01' 
order by salary desc;

9.题目描述
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 t1.dept_no, t1.emp_no, t2.salary 
from dept_manager as t1 
inner join salaries as t2 
on t1.emp_no = t2.emp_no
and t1.to_date = '9999-01-01'
and t2.to_date = '9999-01-01';

10.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));

select emp_no
from employees
where emp_no not in 
(select emp_no from dept_manager);