每天五道MySQL---2

1. 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

2. 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 dept_no,salaries.emp_no,salary
   from salaries left join dept_manager
   on salaries.emp_no = dept_manager.emp_no
   where  dept_manager.to_date ='9999-01-01'
   and salaries.to_date ='9999-01-01'

3. 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 distinct employees.emp_no 
   from employees 
   where employees.emp_no  not in (
   select distinct emp_no from dept_manager
   )

4. 获取所有员工当前的manager，如果当前的manager是自己的话结果不显示，当前表示to_date='9999-01-01'。
   结果第一列给出当前员工的emp_no,第二列给出其manager对应的manager_no。

   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`));
   ---------------------------------
   select dept_emp.emp_no,dept_manager.emp_no as manager_no
   from dept_emp join dept_manager
   where dept_emp.dept_no = dept_manager.dept_no
   and dept_emp.to_date =  dept_manager.to_date
   and dept_emp.emp_no <> dept_manager.emp_no

5. 获取所有部门中当前员工薪水最高的相关信息，给出dept_no, emp_no以及其对应的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 `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 dept_no,salaries.emp_no,Max(salaries.salary) as salary
   from salaries join dept_emp
   on salaries.emp_no = dept_emp.emp_no
   where salaries.to_date = dept_emp.to_date
   and salaries.to_date = '9999-01-01'
   group by dept_no

标签：每天,no,五道,dept,manager,emp,MySQL,date,NULL
来源： https://www.cnblogs.com/luo-bo/p/11134296.html