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
));
思路:
1.先生成两个临时表,1.表示所有员工号、部门号和薪水;2.表示所有管理者员工号、部门号和薪水
2.然后将这两个表相连,条件就是部门号相同,1表的薪水比2表薪水多。
select c.emp_no AS emp_no ,d.emp_no as manager_no , c.salary AS emp_salary , d.salary AS manager_salary
from
(select a.emp_no , b.salary ,a.dept_no
from dept_emp as a join salaries as b
on a.emp_no=b.emp_no
where a.to_date='9999-01-01' and b.to_date='9999-01-01') as c
,
(select a.emp_no ,b.salary,a.dept_no
from dept_manager as a join salaries as b
on a.emp_no=b.emp_no
where a.to_date='9999-01-01' and b.to_date='9999-01-01') as d
where c.dept_no = d.dept_no and c.salary > d.salary;