牛客网sql练习重点

group by 和 max 的结合

• 统计各个部门最高薪水的员工，联结查询分组之后使用max函数。
• 原题

select de.dept_no, de.emp_no, max(s.salary) as salary
from dept_emp de inner join salaries s
on de.emp_no  = s.emp_no
where de.to_date='9999-01-01' and s.to_date='9999-01-01'
group by de.dept_no;

自联结应用

• 从第一张获取入职时候的薪水，从第二张获取当前薪水，从而得到增长。
• 原题

select e.emp_no, (s2.salary - s1.salary) as growth
from employees e 
inner join salaries s1
on e.emp_no = s1.emp_no and s1.from_date = e.hire_date
inner join salaries s2
on e.emp_no = s2.emp_no and s2.to_date = '9999-01-01'
order by growth;

• 对薪水进行排序，第二张表用来统计薪水大于等于待排序薪水的记录的数量。
• 原题

select s1.emp_no, s1.salary, (count(distinct s2.salary)) as rank
from salaries s1 inner join salaries s2
where s1.to_date='9999-01-01' and s2.to_date='9999-01-01' and s1.salary <= s2.salary
group by s1.emp_no
order by s1.salary desc;

• 第一张表示初始薪水，第二张表示结束的薪水。
  
  • 一段时期的开始时间和另一段时期的结束时间在年份上是相同的。
• 原题

select s2.emp_no, s2.from_date, (s2.salary - s1.salary) as salary_growth
from salaries s1 inner join salaries s2
on s1.emp_no = s2.emp_no
and salary_growth > 5000
and (
    strftime('%Y', s2.to_date) - (strftime('%Y', s1.to_date)) = 1
    or strftime('%Y', s2.from_date) - (strftime('%Y', s1.from_date)) = 1
     )
order by salary_growth desc;

• 以第一张表的emp_no分组，在第二张表中通过联结条件计算较小的所有emp_no对应的薪水的和。
• 原题

select s1.emp_no, s1.salary, sum(s2.salary) as running_total
from salaries s1 inner join salaries s2
on s2.emp_no <= s1.emp_no and s2.to_date = '9999-01-01' and s1.to_date = '9999-01-01'
group by s1.emp_no;

从查询的结果中查询

• 查找普通员工当前的薪水和经理当前的薪水，并从这个结果中查找比经理薪水高的员工相关信息。
• 原题

select es.emp_no, ms.emp_no as manager_no, es.salary as emp_salary, ms.salary as manager_salary
from 
(select s.emp_no, de.dept_no, s.salary 
 from salaries s inner join dept_emp de
 on s.emp_no = de.emp_no and s.to_date='9999-01-01') as es,
(select s.emp_no, dm.dept_no, s.salary
 from salaries s inner join dept_manager dm
 on s.emp_no = dm.emp_no and s.to_date='9999-01-01') as ms
where es.dept_no = ms.dept_no and es.salary > ms.salary;

删除重复记录

• 先查找需要保留的id，再将不在范围内的记录删除。
• 原题

delete from titles_test 
where id not in (select min(id) from titles_test group by emp_no);