3

我有一个带有列date和的表time_spent。我想为每个日期 D 找到一段时间内“time_spent”的值的总和：（D-7 - D），即。过去一周 + 当前日期。

我想不出办法来做到这一点，因为我只能找到总和的例子，而不是可变时间段内的总和。

这是一个数据集示例：

CREATE TABLE rolling_total
(
  date date,
  time_spent int
);

INSERT INTO rolling_total VALUES ('2013-09-01','2'),
('2013-09-02','1'),
('2013-09-03','3'),
('2013-09-04','4'),
('2013-09-05','2'),
('2013-09-06','5'),
('2013-09-07','3'),
('2013-09-08','2'),
('2013-09-09','1'),
('2013-09-10','1'),
('2013-09-11','1'),
('2013-09-12','3'),
('2013-09-13','2'),
('2013-09-14','4'),
('2013-09-15','6'),
('2013-09-16','1'),
('2013-09-17','2'),
('2013-09-18','3'),
('2013-09-19','4'),
('2013-09-20','1'),
('2013-09-21','6'),
('2013-09-22','5'),
('2013-09-23','3'),
('2013-09-24','1'),
('2013-09-25','5'),
('2013-09-26','2'),
('2013-09-27','1'),
('2013-09-28','4'),
('2013-09-29','3'),
('2013-09-30','2')

结果看起来像：

date       | time_spent   | rolling_week_total
2013-09-01 |           2  |         2
2013-09-02 |           1  |         3
2013-09-03 |           3  |         6
2013-09-04 |           4  |        10
2013-09-05 |           2  |        12
2013-09-06 |           5  |        17
2013-09-07 |           3  |        20
2013-09-08 |           2  |        22
// now we omit values that are older than seven days
2013-09-09 |           1  |        21
2013-09-10 |           1  |        21
...
4

4 回答 4

8

还有另一种解决方案

SELECT r1.date, r1.time_spent, sum(r2.time_spent) AS rolling_week_total
FROM rolling_total AS r1 JOIN rolling_total AS r2
    ON datediff(r1.date, r2.date) BETWEEN 0 AND 7
GROUP BY r1.date
ORDER BY r1.date
LIMIT 8
于 2013-10-06T21:54:05.557 回答
0

这是另一个解决方案

SELECT r1.date, r1.time_spent, 
   ( SELECT SUM(time_spent) 
     FROM rolling_total AS r2
     WHERE datediff(r1.date, r2.date) 
     BETWEEN 0 AND 7
   ) AS rolling_week_total
FROM rolling_total AS r1  
ORDER BY r1.date
LIMIT 8
于 2013-10-06T21:15:56.170 回答
0

干得好：

select
t.*,
@total:=@total + t.time_spent - coalesce(t2.time_spent, 0) as your_total_over_last_7_days
from
rolling_total t
left join rolling_total t2 on t.date = t2.date + interval 8 day
, (select @total:=0) v
order by t.`date`
于 2013-10-06T21:03:01.710 回答
0

MySQL 8 有专门针对这种情况的窗口函数：

SELECT
    SUM(time_spent) OVER(
         ORDER BY date RANGE BETWEEN INTERVAL 7 DAY PRECEDING 
         AND CURRENT ROW) AS total,
    date
FROM rolling_total

ORDER BY日期确定您用于滚动窗口的维度。

RANGE BETWEEN A AND B定义窗口的过滤条件。

INTERVAL 7 DAY PRECEDING表示当前行之前的所有 7 天。

CURRENT ROW使用当前行的值。

于 2022-01-13T19:25:42.117 回答