根据列的总和返回最小值和最大值

Question

我有一张表Employees，列有：day、employeeid、earned、hourofday。所以我有不同的行，例如：

day        | employeeid | earned | hourofday
25/01/2019 | 422        | 50     | 14
25/01/2019 | 422        | 60     | 15
25/01/2019 | 422        | 40     | 16

我想知道哪个员工每天的最低和最高总收入？

到目前为止，我已经设法返回了总收入/小时数： SELECT day, employeeid, SUM(earned_that_hour) as earned_that_day FROM Employees GROUP BY day, employeeid

然后我尝试在earned_that_day 列上使用 MAX() 和 MIN()：

SELECT day, campaignid, MAX(earned_that_day), MIN(earned_that_day) 
FROM (SELECT day, employeeid, SUM(earned_that_hour) as earned_that_day FROM Employees GROUP BY day, employeeid) 
GROUP BY day, employeeid

但是 MAX 和 MIN 列都返回总和，而不是每天 &employeeid 的最小值和最大值。如何修改查询以返回？我现在有 4 天

Answer 1

使用 2 个 CTE，获取每位员工每天的总和以及每天的最大值/最小值。
然后加入他们：

WITH 
  cte AS (
    SELECT day, employeeid, SUM(earned_that_hour) as earned_that_day 
    FROM Employees 
    GROUP BY day, employeeid
  ),
  maxmin AS (
    SELECT day, MIN(earned_that_day) minearned, MAX(earned_that_day) maxearned
    FROM cte
    GROUP BY day
  ) 
SELECT c.*
FROM cte c inner join maxmin m
ON m.day = c.day AND (c.earned_that_day IN (m.minearned, m.maxearned))
ORDER BY c.day, c.earned_that_day

Answer 2

在较新版本的 SQLite 中，您可以使用窗口函数：

SELECT e.*
FROM (SELECT day, employeeid, SUM(earned_that_hour) as earned_that_day,
             MIN(SUM(earned_that_hour)) OVER (PARTITION BY DAY) as day_min,
             MAX(SUM(earned_that_hour)) OVER (PARTITION BY DAY) as day_max
      FROM Employees e
      GROUP BY day, employeeid
     ) e
WHERE earned_that_day IN (day_min, day_max)
ORDER BY day;