题目描述
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`));
输出描述
| emp_no | salary | from_date | to_date | dept_no |
|---|---|---|---|---|
| 10002 | 72527 | 2001-08-02 | 9999-01-01 | d001 |
| 10004 | 74057 | 2001-11-27 | 9999-01-01 | d004 |
| 10005 | 94692 | 2001-09-09 | 9999-01-01 | d003 |
| 10006 | 43311 | 2001-08-02 | 9999-01-01 | d002 |
| 10010 | 94409 | 2001-11-23 | 9999-01-01 | d006 |
试题详解
分别从两个表中取出要求列并限制员工编号emp_no相同且时间to_date都为当前，之后通过ORDER BY顺序输出。代码如下：
SELECT dm.emp_no,s.salary,s.from_date,dm.to_date,dm.dept_no
FROM dept_manager AS dm,salaries AS s
WHERE dm.emp_no = s.emp_no
AND dm.to_date = '9999-01-01'
AND s.to_date = '9999-01-01'
ORDER BY dm.emp_no;