我当前在数据库中具有以下简化表。积分表包含针对每位用户投票的每种投标形式授予的积分行。
我想在此表中添加一列，该列的每一行显示授予该用户的先前两分的平均值。
Users
+----+----------------------+
| id | name |
+----+----------------------+
| 1 | Flossie Schamberger |
| 2 | Lawson Graham |
| 3 | Hadley Reilly |
+----+----------------------+
Bid Forms
+----+-----------------+
| id | name |
+----+-----------------+
| 1 | Summer 2017 |
| 2 | Winter 2017 |
| 3 | Summer 2018 |
| 4 | Winter 2019 |
| 5 | Summer 2019 |
+----+-----------------+
Points
+-----+---------+--------------------+------------+------------+
| id | user_id | leave_bid_forms_id | bid_points | date |
+-----+---------+--------------------+------------+------------+
| 1 | 1 | 1 | 6 | 2016-06-19 |
| 2 | 2 | 1 | 8 | 2016-06-19 |
| 3 | 3 | 1 | 10 | 2016-06-19 |
| 4 | 1 | 2 | 4 | 2016-12-18 |
| 5 | 2 | 2 | 8 | 2016-12-18 |
| 6 | 3 | 2 | 4 | 2016-12-18 |
| 7 | 1 | 3 | 10 | 2017-06-18 |
| 8 | 2 | 3 | 12 | 2017-06-18 |
| 9 | 3 | 3 | 4 | 2017-06-18 |
| 10 | 1 | 4 | 4 | 2017-12-17 |
| 11 | 2 | 4 | 4 | 2017-12-17 |
| 12 | 3 | 4 | 2 | 2017-12-17 |
| 13 | 1 | 5 | 16 | 2018-06-17 |
| 14 | 2 | 5 | 12 | 2018-06-17 |
| 15 | 3 | 5 | 10 | 2018-06-17 |
+-----+---------+--------------------+------------+------------+
对于点表中的每一行，我希望按如下方式计算average_points列。
“平均分”列是该用户之前2个点的平均值。因此，对于该表中每个用户的第一个条目，平均值显然为0，因为之前没有授予任何积分。
应使用日期列确定每个用户的前2点。
下表是我希望作为最终输出的内容。 为了清楚起见，在表的一边，我添加了用于得出averaged_points列中值的计算和数字。
+-----+---------+--------------------+------------+-----------------+
| id | user_id | leave_bid_forms_id | date | averaged_points |
+-----+---------+--------------------+------------+-----------------+
| 1 | 1 | 1 | 2016-06-19 | 0 | ( 0 + 0 ) / 2
| 2 | 2 | 1 | 2016-06-19 | 0 | ( 0 + 0 ) / 2
| 3 | 3 | 1 | 2016-06-19 | 0 | ( 0 + 0 ) / 2
| 4 | 1 | 2 | 2016-12-18 | 3 | ( 6 + 0 ) / 2
| 5 | 2 | 2 | 2016-12-18 | 4 | ( 8 + 0 ) / 2
| 6 | 3 | 2 | 2016-12-18 | 5 | ( 10 + 0) / 2
| 7 | 1 | 3 | 2017-06-18 | 5 | ( 4 + 6 ) / 2
| 8 | 2 | 3 | 2017-06-18 | 8 | ( 8 + 8 ) / 2
| 9 | 3 | 3 | 2017-06-18 | 7 | ( 4 + 10) / 2
| 10 | 1 | 4 | 2017-12-17 | 7 | ( 10 + 4) / 2
| 11 | 2 | 4 | 2017-12-17 | 10 | ( 12 + 8) / 2
| 12 | 3 | 4 | 2017-12-17 | 4 | ( 4 + 4 ) / 2
| 13 | 1 | 5 | 2018-06-17 | 7 | ( 4 + 10) / 2
| 14 | 2 | 5 | 2018-06-17 | 8 | ( 4 + 12) / 2
| 15 | 3 | 5 | 2018-06-17 | 3 | ( 2 + 4 ) / 2
+-----+---------+--------------------+------------+-----------------+
我一直在尝试使用子查询来解决此问题，因为AVG似乎不受我拥有的任何LIMIT子句的影响。
到目前为止，我已经想到了
select id, user_id, leave_bid_forms_id, `date`,
(
SELECT
AVG(bid_points)
FROM (
Select `bid_points`
FROM points as p2
ORDER BY p2.date DESC
Limit 2
) as thing
) AS average_points
from points as p1
这是in this sqlfiddle，但老实说，我在这里没什么意思。
我在正确的道路上吗？想知道是否有人可以告诉我我需要调整的地方！
谢谢。
编辑
使用以下答案作为基础，我能够调整sql以使其与原始sqlfiddle中提供的表一起使用。
我已将其添加到this sqlfiddle to show it working
与上面的代码匹配的更正的sql是
select p.*,
IFNULL(( (coalesce(points_1, 0) + coalesce(points_2, 0)) /
( (points_1 is not null) + (points_2 is not null) )
),0) as prev_2_avg
from (select p.*,
(select p2.bid_points
from points p2
where p2.user_id = p.user_id and
p2.date < p.date
order by p2.date desc
limit 1
) as points_1,
(select p2.bid_points
from points p2
where p2.user_id = p.user_id and
p2.date < p.date
order by p2.date desc
limit 1, 1
) as points_2
from points as p
) p;
尽管我将要问另一个问题，以最佳的方式使这种动态变化与需要平均的以前的点数保持一致。
答案 0 :(得分：1)
您可以使用MySQL 8中引入的window functions。
select p.*,
avg(points) over (partition by user_id
order by date
rows between 2 preceding and 1 preceding
) as prev_2_avg
from p;
在早期版本中，这是一个真正的痛苦，因为MySQL不支持嵌套的关联子句。一种方法是每个方法都有一个单独的列：
select p.*,
( (coalesce(points_1, 0) + coalesce(points_2, 0)) /
( (points_1 is not null) + (points_2 is not null) )
) as prev_2_avg
from (select p.*,
(select p2.points
from points p2
where p2.user_id = p.user_id and
p2.date < p.date
order by p2.date desc
limit 1
) as points_1,
(select p2.points
from points p2
where p2.user_id = p.user_id and
p2.date < p.date
order by p2.date desc
limit 1, 1
) as points_2
from p
) p;