获取连续2年获得加薪的员工

Question

我正在尝试获取连续 2 年获得加薪的员工，在这种情况下，员工 1000 是正确答案。

这是我尝试过的数据和sql。

EID	SALARY	YEAR
1000	10,000	2016
1000	7,500	2015
1000	6,000	2014
1001	8,000	2016
1001	7,500	2015
1002	7,500	2016
1002	7,500	2015
1002	5,000	2014
1003	6,000	2016
1003	7,000	2015
1003	5,000	2014

我使用了以下代码，但是它仅按 EID 获取行号，而不是去年和今年的计算，我需要找到连续 2 年加薪的员工。

output

select * , 
       row_number() over(partition by eid order by salary and year desc)as rn 
from gs;

Answer 1

您可以使用LEAD 窗口函数来计算工资的两个连续前值。然后您可以使用salary1 < salary2 < salary3 检查有多少员工至少有一行。

SELECT DISTINCT 
    eid
FROM (
    SELECT
        eid,
        year,
        salary,
        (LEAD(salary, 1) OVER(PARTITION BY eid ORDER BY year DESC)) AS prev_salary1,
        (LEAD(salary, 2) OVER(PARTITION BY eid ORDER BY year DESC)) AS prev_salary2
    FROM 
        employees
    ) consecutive3
WHERE 
    salary > prev_salary1
AND     
    prev_salary1 > prev_salary2

假设没有缺失的年份没有记录受抚养人的工资。

这是小提琴：https://dbfiddle.uk/?rdbms=mysql_8.0&fiddle=8c0d8a1deec8e77bb32a173656c3e386。

---

编辑：详细解释

让我们以詹妮弗为例，她工作了五年，拿到了这些薪水：

• 2018 -> 65000
• 2017 -> 55000
• 2016 -> 50000

她是被选上的候选人，因为她的薪水连续提高了三倍。

1.LEAD(salary, 1) OVER(PARTITION BY eid ORDER BY year DESC)

允许您获取“Y”年的薪水和“Y-1”年的薪水：

（“年份”->“薪水”，“previous_salary”）

• 2018 -> 65000 , 55000
• 2017 -> 55000 , 50000
• 2016 -> 50000，空

2.LEAD(salary, 2) OVER(PARTITION BY eid ORDER BY year DESC)

允许您获取“Y”年的薪水和“Y-1”年的薪水：

（“年”->“薪水”，“previous_salary”，“previous_salary_by_2_years”）

• 2018 -> 65000 , 55000 , 50000
• 2017 -> 55000 , 50000 , 空
• 2016 -> 50000，空，空

3.WHERE salary > prev_salary1 AND prev_salary1 > prev_salary2

对员工进行一些过滤

• 第三年的薪水高于第二年的薪水 (salary > prev_salary1)
• 他们第二年的薪水高于第一年的薪水 (prev_salary1 > prev_salary2)

Answer 2

我知道这已经得到解答，但这是我使用滞后函数确定是否比前一年增加并运行两次的结果。

SELECT *
FROM (
    SELECT
        t2.*,
        LAG(increase) over (partition by eid order by year) AS increaseNextYear
    FROM (
        SELECT
            t1.*,
            COALESCE(salary - LAG(salary) over (partition by eid order by year), 0) > 0 AS increase
        FROM tbl_test t1
    ) as t2
) t3 where increase AND increaseNextYear

Answer 3

with
  dates as
(
  select
    a.*,
    dense_rank() OVER (
      partition by eid
          order by year desc, salary
      )
        as rn, 
    case
      when 
        lead(salary,2)over(partition  by eid order by year, salary)
        >
        lead(salary,1)over(partition by eid order by year, salary)
      and  
        lead(salary,1)over(partition by eid order by year, salary)
        >
        salary
      then
        1
      else
        0
    end
      as flag
  from
    employees a
)
select
  eid
from
  dates
where
      rn   = 3
  and flag = 1

Answer 4

不是一个漂亮的查询，而是直截了当的：查找在上一年的薪水较低且前一年的薪水更低的一年的员工。使用LAG 更优雅，但我想我会把它扔进去，只是为了展示一个替代方案。

select *
from employee
where exists
(
  select null
  from gs
  where gs.eid = employee.id
  and exists
  (
    select null
    from gs prev
    where prev.eid = gs.eid
    and prev.year = gs.year - 1
    and prev.salary < gs.salary
    and exists
    (
      select null
      from gs prevprev
      where prevprev.eid = prev.eid
      and prevprev.year = prev.year - 1
      and prevprev.salary < prev.salary
    )
  )
);

加入同样的事情：

select *
from employee
where exists
(
  select null
  from gs
  join gs prev on prev.eid = gs.eid
              and prev.year = gs.year - 1
              and prev.salary < gs.salary
  join gs prevprev on prevprev.eid = prev.eid
                  and prevprev.year = prev.year - 1
                  and prevprev.salary < prev.salary
  where gs.eid = employee.id
);

Answer 5

对于 8.0 之前的版本（我的是 5.7），缺少新版本的尖端功能，我尝试了一个程序来完成这项工作。首先，获得所有开斋节的工资记录不少于三年，这是连续奖金的最低要求。剩下的就是使用 eid 池中的游标来获取和比较。结果存储在临时表 t 中。

delimiter //
drop procedure if exists lucky_emp//
create procedure lucky_emp()
begin
declare old_eid int default 0;
declare old_salary int;
declare new_eid int ;
declare new_salary int;
declare bonus_year int;
declare fin bool default false;
declare c cursor for select eid,salary from salary where eid in(select eid from salary group by eid having count(eid)>=3) order by eid,year;
declare continue handler for not found set fin=true;
drop temporary table if exists t ;
create temporary table t (t_eid int);
open c;
lp:loop
    fetch c into new_eid ,new_salary;
    if fin=true then
        leave lp;
    end if;
    if new_eid !=old_eid then
        set old_eid=new_eid,old_salary=0,bonus_year=0;
    end if;
    if new_salary> old_salary then
        set bonus_year=bonus_year+1,old_salary=new_salary;
    else 
        set bonus_year=0;
    end if;

    if bonus_year=3 and new_eid not in(select t_eid from t) then
        insert t values(new_eid);
    end if;

end loop lp;

end// 
delimiter ;
select * from t ;

Answer 6

Select a.*, b.prev_sal1, b.prev_sal2
from employees a
join (
     Select eid ,year,
     lag(salary,1) over (partition by eid order by year) as prev_sal1,
     lag(salary,2) over (partition by eid order by year) as prev_sal2
     from employees ) b
on a.eid=b.eid
and a.year = b.year
where salary>prev_sal1 and prev_sal1>prev_sal2

小提琴：https://dbfiddle.uk/rfGv31zM