我有一个类似于以下表格：
场地
下表描述了企业列表
id name
50 Nando's
60 KFC
奖励
该表描述了许多奖励，奖励所对应的地点以及兑换奖励所需的积分。
id venue_id name points
1 50 5% off 10
2 50 10% off 20
3 50 11% off 30
4 50 15% off 40
5 50 20% off 50
6 50 30% off 50
7 60 30% off 70
8 60 60% off 100
9 60 65% off 120
10 60 70% off 130
11 60 80% off 140
points_data
该表描述了用户在每个场所剩余的积分数。
venue_id points_remaining
50 30
60 90
请注意，此查询实际上是在SQL中计算的，如下所示：
select * from (
select venue_id, (total_points - points_redeemed) as points_remaining
from (
select venue_id, sum(total_points) as total_points, sum(points_redeemed) as points_redeemed
from (
(
select venue_id, sum(points) as total_points, 0 as points_redeemed
from check_ins
group by venue_id
)
UNION
(
select venue_id, 0 as total_points, sum(points) as points_redeemed
from reward_redemptions rr
join rewards r on rr.reward_id = r.id
group by venue_id
)
) a
group by venue_id
) b
GROUP BY venue_id
) points_data
但是对于这个问题，您可能可以忽略该大量查询，并假设该表仅称为points_data。
我想得到一个查询，得到：
因此对于上述数据，输出为：
id venue_id name points
2 50 10% off 20
3 50 11% off 30
4 50 15% off 40
5 50 20% off 50
7 60 30% off 70
8 60 60% off 100
9 60 65% off 120
到目前为止，我发现的最佳解决方案是首先获取points_data，然后使用代码（即PHP）动态编写以下内容：
(
select * from rewards
where venue_id = 50
and points > 30
ORDER BY points desc
LIMIT 2
)
union all
(
select * from rewards
where venue_id = 50
and points <= 30
ORDER BY points desc
LIMIT 2
)
UNION ALL
(
select * from rewards
where venue_id = 60
and points <= 90
ORDER BY points desc
LIMIT 2
)
UNION ALL
(
select * from rewards
where venue_id = 60
and points > 90
ORDER BY points desc
LIMIT 2
)
ORDER BY venue_id, points asc;
但是，我觉得查询可能会变得太长且效率低下。例如，如果用户在400个场所中拥有积分，则为800个子查询。
我也尝试过这样的联接，但是并不能真正做到更好：
select * from points_data
INNER JOIN rewards on rewards.venue_id = points_data.venue_id
where points > points_remaining;
这远不是我想要的。
答案 0 :(得分：2)
相关子查询计算较高或较低奖励的数量来确定顶部或底部条目是一种方法。
SELECT r1.*
FROM rewards r1
INNER JOIN points_data pd1
ON pd1.venue_id = r1.venue_id
WHERE r1.points <= pd1.points_remaining
AND (SELECT count(*)
FROM rewards r2
WHERE r2.venue_id = r1.venue_id
AND r2.points <= pd1.points_remaining
AND (r2.points > r1.points
OR r2.points = r1.points
AND r2.id > r1.id)) < 2
OR r1.points > pd1.points_remaining
AND (SELECT count(*)
FROM rewards r2
WHERE r2.venue_id = r1.venue_id
AND r2.points > pd1.points_remaining
AND (r2.points < r1.points
OR r2.points = r1.points
AND r2.id < r1.id)) < 2
ORDER BY r1.venue_id,
r1.points;
自MySQL 8.0起，使用row_number()窗口函数的解决方案将是替代方案。但我想您的版本较低。
SELECT x.id,
x.venue_id,
x.name,
x.points
FROM (SELECT r.id,
r.venue_id,
r.name,
r.points,
pd.points_remaining,
row_number() OVER (PARTITION BY r.venue_id,
r.points <= pd.points_remaining
ORDER BY r.points DESC) rntop,
row_number() OVER (PARTITION BY r.venue_id,
r.points > pd.points_remaining
ORDER BY r.points ASC) rnbottom
FROM rewards r
INNER JOIN points_data pd
ON pd.venue_id = r.venue_id) x
WHERE x.points <= x.points_remaining
AND x.rntop <= 2
OR x.points > x.points_remaining
AND x.rnbottom <= 2
ORDER BY x.venue_id,
x.points;
这里最棘手的部分是将每个地点的集合也划分为子集，在该子集中，用户的积分足以兑现奖励；在子集中，积分不足。但是，就像在MySQL逻辑表达式中计算为0或1（在非布尔上下文中）一样，相应的表达式也可以用于此。