SQL查询以查找加薪3年的员工？

这是桌子。我最初的观察是询问员工的年薪增长情况，但我不知道如何做到这一点。员工1是唯一一个有三年加薪的员工，但不知道如何挑选他们。谢谢

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一个选项使用聚合：

select employee_id
from mytable t
group by employee_id
having max(salary) filter(where year = 2020) > max(salary) filter(where year = 2019)
   and max(salary) filter(where year = 2019) > max(salary) filter(where year = 2018)

select employee_id
from (select t.*,
             lag(salary) over (partition by employee_id order by year) as prev_salary
      from t
     ) t
group by employee_id
having min(salary - prev_salary) > 0 and
       count(*) = 3;

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这就带来了2020年工资高于2019年工资、2019年工资高于2018年工资的员工——这就是我对你问题的理解。

你可以使用

lead（）

/

lag（）

和聚合：

select employee_id
from mytable t
group by employee_id
having max(salary) filter(where year = 2020) > max(salary) filter(where year = 2019)
   and max(salary) filter(where year = 2019) > max(salary) filter(where year = 2018)

select employee_id
from (select t.*,
             lag(salary) over (partition by employee_id order by year) as prev_salary
      from t
     ) t
group by employee_id
having min(salary - prev_salary) > 0 and
       count(*) = 3;

这将比较相邻年份的工资，并返回价值不断增加的员工。它假设这些年中没有间隔——就像在您的样本数据中一样

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这种方法的优点是您不需要提前知道年份。

我想您不想在查询中硬编码年份。
最佳选择是使用窗口函数

LAG（）

获取前2年的工资，但您还应检查您检查的3年是否连续：

SELECT DISTINCT employee_id
FROM (
  SELECT *, 
    LAG(year, 1) OVER (PARTITION BY employee_id ORDER BY year) year1,
    LAG(salary, 1) OVER (PARTITION BY employee_id ORDER BY year) salary1,
    LAG(year, 2) OVER (PARTITION BY employee_id ORDER BY year) year2,
    LAG(salary, 2) OVER (PARTITION BY employee_id ORDER BY year) salary2 
  FROM tablename t
)
WHERE year1 = year - 1 AND year2 = year - 2 AND salary > salary1 AND salary1 > salary2 

如果只想检查当前年份和前两年，则在

WHERE

子句中再添加一个条件：

...AND year = strftime('%Y', CURRENT_DATE) 

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因此，您不需要硬编码当前年份，

是的，这很有效！这使得我每年把工资加起来更有意义。谢谢你！