数据库SQL实战-- 获取员工其当前的薪水比其manager当前薪水还高的相关信息

题目描述
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));

解题思路：

因为需要员工工资比经理高的数据，所以需要两张表，
第一张是所有员工的工资，第二章是经理的工资
然后根据条件比较

SQL:
第一种写法： 耗时53ms
select s1.emp_no,s2.emp_no,s1.salary,s2.salary
from(select p.emp_no,p.dept_no,s.salary
from dept_emp p inner join salaries s
on p.emp_no = s.emp_no and p.to_date='9999-01-01' and s.to_date='9999-01-01'
) as s1 inner join
(
select r.emp_no,r.dept_no,s.salary
from dept_manager r inner join salaries s
on r.emp_no = s.emp_no and r.to_date='9999-01-01' and s.to_date='9999-01-01'
) as s2
on s1.dept_no = s2.dept_no where s1.salary >s2.salary

第二种写法： 耗时23ms

select de.emp_no ,(dm.emp_no) as manager_no , (s1.salary) as emp_salary ,(s2.salary) as manager_salary
from salaries as s1, salaries as s2
inner join dept_emp as de on de.emp_no =s1.emp_no
inner join dept_manager as dm on dm.emp_no =s2.emp_no
where de.dept_no=dm.dept_no
and s1.to_date='9999-01-01'
and s2.to_date='9999-01-01'
and s1.salary>s2.salary