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。
SELECT * FROM employees ORDER BY hire_date DESC LIMIT 0,1;但是有人说最晚入职不一定只有一个人，所以选择最大日期即可：
select * from employees
where hire_date =
(select max(hire_date) from employees)[编程题]查找入职员工时间排名倒数第三的员工所有信息
select * from employees order by hire_date desc limit 2,1;limit后第一个2为no_of_rows即第几行，一般如果没有后面的 1 ，就会返回从0到2共3行，加了1就是从2开始后的一行，1即offset偏移量；
[编程题]查找当前薪水详情以及部门编号dept_no
SELECT s.*,d.dept_no
FROM salaries s inner join dept_manager d on s.emp_no = d.emp_no
WHERE s.to_date='9999-01-01'
AND d.to_date='9999-01-01';说实话我觉得这两个表有问题，emp_no一致怎么会有不同的to_date?违反了完整性约束啊；
[编程题]查找所有已经分配部门的员工的last_name和first_name
select e.last_name,e.first_name,d.dept_no from employees e cross join dept_emp d on e.emp_no = d.emp_no;[编程题]查找所有员工的last_name和first_name以及对应部门编号dept_no
SELECT e.last_name, e.first_name, d.dept_no
FROM employees e
LEFT JOIN dept_emp d
ON e.emp_no = d.emp_no;编程题]查找所有员工入职时候的薪水情况
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;链接：https://www.nowcoder.com/questionTerminal/23142e7a23e4480781a3b978b5e0f33a
来源：牛客网
1、由于测试数据中，salaries.emp_no 不唯一（因为号码为 emp_no 的员工会有多次涨薪的可能，所以在 salaries 中对应的记录不止一条），employees.emp_no 唯一，即 salaries 的数据会多于 employees，因此需先找到 employees.emp_no 在 salaries 表中对应的记录salaries.emp_no，则有限制条件 e.emp_no = s.emp_no
2、根据题意注意到 salaries.from_date 和 employees.hire_date 的值应该要相等，因此有限制条件 e.hire_date = s.from_date
[编程题]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
select emp_no,count(*) as t from salaries group by emp_no having count(*) > 15;[编程题]找出所有员工当前薪水salary情况
select distinct salary from salaries where to_date = '9999-01-01' order by salary desc;[编程题]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'
select d.dept_no,s.emp_no,s.salary from dept_manager d inner join salaries s on d.emp_no = s.emp_no
where d.to_date = '9999-01-01'
group by d.dept_no;[编程题]获取所有非manager的员工emp_no
select emp_no from employees where emp_no not in (select emp_no from dept_manager);[编程题]获取所有员工当前的manager
这个我写了两种：
select e.emp_no,m.emp_no as manager_no
from dept_emp e inner join dept_manager m on e.dept_no = m.dept_no
where e.emp_no not in (select emp_no from dept_manager) and e.to_date='9999-01-01' AND m.to_date = '9999-01-01';SELECT de.emp_no, dm.emp_no AS manager_no
FROM dept_emp AS de INNER JOIN dept_manager AS dm
ON de.dept_no = dm.dept_no
WHERE dm.to_date = '9999-01-01' AND de.to_date = '9999-01-01' AND de.emp_no <> dm.emp_no第二种好一点，因为不等于更具普遍性
[编程题]获取所有部门中当前员工薪水最高的相关信息
select d.dept_no,d.emp_no,max(s.salary) from dept_emp as d inner join salaries as s on d.emp_no=s.emp_no
WHERE d.to_date = '9999-01-01' AND s.to_date = '9999-01-01'
group by d.dept_no;[编程题]从titles表获取按照title进行分组
select title,count(title) as t from titles group by title having t >= 2;[编程题]从titles表获取按照title进行分组，注意对于重复的emp_no进行忽略。
这题很不错
https://www.nowcoder.com/practice/c59b452f420c47f48d9c86d69efdff20?tpId=82&tqId=29766&tPage=
1&rp=&ru=%2Fta%2Fsql&qru=%2Fta%2Fsql%2Fquestion-ranking
1、先用GROUP BY title将表格以title分组，再用COUNT(DISTINCT emp_no)可以统计同一title值且不包含重复emp_no值的记录条数
select distinct title,count(distinct emp_no) as t from titles group by title having count(title)>=2;[编程题]查找employees表
select * from employees where emp_no%2 = 1 and last_name != 'Mary' order by hire_date desc;[编程题]统计出当前各个title类型对应的员工当前薪水对应的平均工资
select t.title,avg(s.salary) as avg
from salaries s inner join titles t on s.emp_no = t.emp_no
where t.to_date='9999-01-01'
and s.to_date='9999-01-01'
group by t.title;[编程题]获取当前薪水第二多的员工的emp_no以及其对应的薪水salary
select emp_no,salary
from salaries
where to_date = '9999-01-01'
order by salary desc limit 1,1;考虑到钱多的人可能不止一个，所以更好的语句是：
select emp_no, salary
from salaries
where salary = (select salary from salaries group by salary order by salary desc limit 1,1)[编程题]获取当前薪水第二多的员工的emp_no以及其对应的薪水salary，不准使用order by
select e.emp_no,s.salary,e.last_name,e.first_name from
employees e inner join salaries s on e.emp_no = s.emp_no
where s.to_date = '9999-01-01' and s.emp_no=(select s1.emp_no from salaries s1 join salaries s2 on s1.salary<=s2.salary
and s1.to_date='9999-01-01' and s2.to_date='9999-01-01'
group by s1.emp_no having count(s1.salary) = 2);这题就很不错，我也不想用max，因为你用了max的局限很明显，如果我想找第三多，第四多的工资员工就会很乏力；
用max的解法
select e.emp_no,s.salary,e.last_name,e.first_name from
employees e inner join salaries s on e.emp_no = s.emp_no
where s.to_date = '9999-01-01'
and
s.salary = (select max(salary) from salaries where salary != (select max(salary) from salaries));[编程题]获取员工其当前的薪水比其manager当前薪水还高的相关信息
select x.emp_no,manager.emp_no as manager_no,x.salary as emp_salary,manager.salary as manager_salary
from
(dept_emp d inner join salaries s on d.emp_no = s.emp_no and d.to_date = '9999-01-01' and s.to_date = '9999-01-01') x
inner join
(dept_manager d inner join salaries s1 on d.emp_no = s1.emp_no and d.to_date = '9999-01-01' and s1.to_date = '9999-01-01')as manager on x.dept_no = manager.dept_no
and x.salary > manager.salary;