獲取員工其當前的薪水比其manager當前薪水還高的相關信息,當前表示to_date=‘9999-01-01’,
結果第一列給出員工的emp_no,
第二列給出其manager的manager_no,
第三列給出該員工當前的薪水emp_salary,
第四列給該員工對應的manager當前的薪水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.先生成兩個臨時表,1.表示所有員工號、部門號和薪水;2.表示所有管理者員工號、部門號和薪水
2.然後將這兩個表相連,條件就是部門號相同,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;