SQL训练汇总（每天更新10道题）
1.
查找最晚入职员工的所有信息
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`));
solution1:(排序 + 限制)
select*from employees order by hire_date desc limit 0, 1;
solution2: (子查询)
select*from employees where hire_date = (select Max(hire_date) from employees);
2.
题目描述
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`));
solution1: (排序 + 限制)
select*from employees order by hire_date desc limit 2, 1;
solution2: (子查询)-主要为了避免同一时间有多个员工报到（排序时去重）
select*from employees where hire_date = (select distinct hire_date from employees order by hire_date desc limit 2, 1);
3.
题目描述
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 s.*, d.dept_no FROM salaries s , dept_manager d
WHERE s.to_date='9999-01-01'
AND d.to_date='9999-01-01'
AND s.emp_no = d.emp_no;
4.
查找所有已经分配部门的员工的last_name和first_name
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 `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`));
solution1:
select last_name, first_name, dept_no from employees e, dept_emp d where d.emp_no = e.emp_no;
5.
查找所有员工的last_name和first_name以及对应部门编号dept_no，也包括展示没有分配具体部门的员工
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 `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`));
solution1:(左连接)
SELECT ep.last_name, ep.first_name, dp.dept_no
FROM employees ep
LEFT JOIN dept_emp dp
ON ep.emp_no = dp.emp_no
注意：
INNER JOIN 两边表同时有对应的数据，即任何一边缺失数据就不显示。
LEFT JOIN 会读取左边数据表的全部数据，即便右边表无对应数据。
RIGHT JOIN 会读取右边数据表的全部数据，即便左边表无对应数据。