牛客SQL题目

题目链接：https://www.nowcoder.com/ta/sql

查找最晚入职员工的所有信息

select * 
from 
employees
where 
hire_date = (select max(hire_date) from employees);

查找入职员工时间排名倒数第3的员工所有信息

select * 
from 
employees 
order by 
hire_date 
desc --降序
limit 2,1;--从3项，偏移1项，即第3项

查找各个部门当前(to_date='9999-01-01')的领导的当前薪水详情以及其对应部门编号dept_no

select s.*,d.dept_no
from salaries as s
join dept_manager as d
on d.emp_no=s.emp_no
where s.to_date='9999-01-01'
and d.to_date='9999-01-01';

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 as e
join dept_emp as d
on e.emp_no = d.emp_no;

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

select e.last_name,e.first_name,d.dept_no
from employees as e
left join dept_emp as d--坐表为主，右表补充，为空补NULL
on e.emp_no = d.emp_no;

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

select e.emp_no,s.salary
from salaries as s
join employees as e
where 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

count( )：统计记录的条数

还需要group by emp_no将每个员工的记录显示在一条记录中
```
select emp_no,count(emp_no) as t
from salaries
group by
emp_no
having t > 15;
```
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

distinct：去除重复
```
select distinct salary-- 去除重复
from salaries
where to_date = '9999-01-01'
order by
salary
desc;
```

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,d.emp_no,s.salary
from dept_manager as d
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';

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);

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'。

select e.emp_no,m.emp_no as manager_no
from dept_emp as e
join dept_manager as m
on e.dept_no=m.dept_no
where e.emp_no <> m.emp_no -- 当前manager是自己不显示，<>不等于
and e.to_date='9999-01-01'
and m.to_date='9999-01-01';

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

select d.dept_no,d.emp_no,max(s.salary)
from dept_emp as d
join salaries as s
on d.emp_no = s.emp_no
group by
d.dept_no;

从titles表获取按照title进行分组，每组个数大于等于2，给出title以及对应的数目t。
```
select title,count(title) as t
from titles
group by
title
having t >= 2;
```
查找employees表所有emp_no为奇数，且last_name不为Mary的员工信息，并按照hire_date逆序排列
```
select * from employees
where emp_no%2 =1
and last_name <> 'Mary'
order by
hire_date
desc;
```

统计出当前各个title类型对应的员工当前（to_date='9999-01-01'）薪水对应的平均工资。结果给出title以及平均工资avg。

select t.title,avg(s.salary)
from titles as t
join salaries as s
on t.emp_no = s.emp_no
where s.to_date='9999-01-01'
and t.to_date='9999-01-01'
group by
t.title;

获取当前（to_date='9999-01-01'）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;  -- 降序，取第二个

查找当前薪水(to_date='9999-01-01')排名第二多的员工编号emp_no、薪水salary、last_name以及first_name，不准使用order by

select e.emp_no,s.salary,e.last_name,e.first_name
from employees as e
join salaries as s
on e.emp_no=s.emp_no
where s.to_date='9999-01-01'
order by
s.salary
desc
limit 1,1;

不使用order by：需要用嵌套select和max结合

select e.emp_no,max(salary) ,e.last_name,e.first_name --这里用max(salary)
from  salaries as s,employees as e
where s.emp_no = e.emp_no
and salary<  -- 小于最大薪水中的最大，就是第二大
(
select max(salary)
from salaries
where to_date = '9999-01-01'
);

查找所有员工的last_name和first_name以及对应的dept_name，也包括暂时没有分配部门的员工

双左连接

select e.last_name,e.first_name,de.dept_name
from employees as e
left join dept_emp as d
on e.emp_no=d.emp_no
left join departments as de
on d.dept_no=de.dept_no;

查找员工编号emp_no为10001其自入职以来的薪水salary涨幅值growth

通过入职时间来排序查询：

如果直接用薪水最大值-最小值，有可能最后一次是降薪。

SELECT ( 
(SELECT salary FROM salaries WHERE emp_no = 10001 ORDER BY to_date DESC LIMIT 1) -
(SELECT salary FROM salaries WHERE emp_no = 10001 ORDER BY to_date ASC LIMIT 1)
) AS growth

查找所有员工自入职以来的薪水涨幅情况，给出员工编号emp_no以及其对应的薪水涨幅growth，并按照growth进行升序

--将两个查询的结果，当成两个表，进行join
select S2.emp_no, (S2.salary-S1.salary) as growth
from 
(select e.emp_no,s.salary 
 from employees as e 
 join salaries as s 
 on e.emp_no=s.emp_no 
 and e.hire_date=s.from_date) as S1  --所有员工的入职工资
join 
(select e.emp_no,s.salary 
 from employees as e 
 join salaries as s 
 on e.emp_no=s.emp_no 
 and s.to_date='9999-01-01') as S2   --所有员工的当前工资
on S1.emp_no = S2.emp_no
order by growth;

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