查找最晚入职员工的所有信息
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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`));
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select * from employees where hire_date= (select max(hire_date) from employees)
需要注意的是,因为最晚入职的人可能不止一人,所以用top不行。
查找入职员工时间排名倒数第三的员工所有信息
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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`));
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select * from employees where hire_date =( select hire_date from employees order by hire_date desc limit 2,1)
查找当前薪水详情以及部门编号dept_no
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查找各个部门当前(to_date='9999-01-01')领导当前薪水详情以及其对应部门编号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`));
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select b.*, a.dept_no from salaries as b inner join dept_manager as a on a.emp_no=b.emp_no where b.to_date='9999-01-01' and a.to_date='9999-01-01'
查找所有员工入职时候的薪水情况
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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`));
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select e.emp_no, s.salary from employees as e inner join salaries as s on e.emp_no = s.emp_no and e.hire_date = s.from_date order by e.emp_no desc
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
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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`));
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select emp_no, count(*) as t from salaries group by emp_no having t>15
找出所有员工当前薪水salary情况
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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`));
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select salary from salaries where to_date='9999-01-01' group by salary order by salary desc
获取所有部门当前manager的当前薪水情况
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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`));
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select d.dept_no, d.emp_no, s.salary from dept_manager as d, salaries as s where d.emp_no = s.emp_no and d.to_date = '9999-01-01' and s.to_date = '9999-01-01'
获取所有非manager的员工emp_no
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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`));
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select emp_no from employees where emp_no not in( select emp_no from dept_manager ) select e.emp_no from employees as e left join dept_manager as d on e.emp_no = d.emp_no where d.emp_no is null
获取所有员工当前的manager
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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`));
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select d1.emp_no, d2.emp_no as manager_no from dept_emp as d1 inner join dept_manager as d2 on d1.dept_no = d2.dept_no where d1.emp_no <> d2.emp_no and d2.to_date='9999-01-01'