Postgres：根据结果加入两个查询和 select

Question

有一个社交网络，每个用户都可以转发另一个用户的帖子。每转发 10 次您的帖子，您将获得一份礼物。有两个表：gifts 和 repost_history，请参见下面的方案。

问题： 如何编写一个查询来计算我需要为系统中的每个用户赠送多少礼物？

=========
= gifts =
=========
   id       // PK
 user_id    // id of user which received a gift
 amount     // amount of gifts (bonuses), may be + or -
  type      // type of a gift.  The only type we're interested in is 'REPOST_TYPE'

==================
= repost_history =
==================
      id     // PK
    user_id  // id of a user which did repost
    owner_id // id of a user whose post was reposted

查询算法：

1) 查找每个用户的总转发次数 SELECT owner_id, COUNT(owner_id) FROM repost_history GROUP BY owner_id;

2)求出每位用户的REPOST_TYPE礼物总金额 SELECT user_id, COUNT(amount) FROM gifts WHERE type = 'REPOST_TYPE' GROUP BY user_id;

3) 根据owner_id = user_id

加入第一步和第二步

4) From (user_id, gift_to_grand_count) 基于第3步结果的结果集。其中 <gift_to_grand_count> = (<reposts_of_user> / 10) - <user_repost_gifts_amount>

我的解决方法： 1-3 步实施（不起作用，因为我不知道如何将子查询结果设置为变量）。如何让它工作并执行第 4 步？

(
  SELECT owner_id, COUNT(owner_id) AS reposts_count
  FROM reposts_history
  GROUP BY owner_id 
  AS user_reposts
)   
INNER JOIN (
  SELECT user_id, COUNT(amount) AS gifts_count
  FROM gifts 
  WHERE type = 'REPOST_GIFT' 
  GROUP BY user_id 
  AS user_gifts
) 
ON user_reposts.owner_id = user_gifts.user_id

数据样本：

为简单起见，假设我们想在每 3 次转发（而不是每 10 次）时赠送一份礼物

gifts - 你可以看到 user_id=1 已获得 1 份 REPOST_TYPE 的礼物。我们对他花了多少礼物不感兴趣。

id | user_id | amount |     type      |
 1 |    1    |    1   | 'REPOST_TYPE' |
 2 |    1    |    2   | 'OTHER_TYPE'  |
 3 |    1    |   -1   | 'REPOST_TYPE' |
 4 |    2    |    1   | 'REPOST_TYPE' |

reposts_history - 您可以看到用户 owner_id=1 被其他用户转发了 6 次。

id  | user_id | owner_id | another columns...
 1  |    2    |    1     |
 2  |    3    |    1     |
 3  |    4    |    1     |
 4  |    5    |    1     |
 5  |    2    |    1     |
 6  |    6    |    1     |
 6  |   13    |    2     |

所以 user_id=1 应该被授予 <total_reposts> / 3 - <already_granted_gifts_amount> = 6 / 3 - 1 = 1 礼物。

我想为系统中的所有用户获取：

user_id | gifts_to_grant |
   1    |       1        |
   2    |       0        |
     ..........

Answer 1

您需要外部联接才能同时找到应得礼物但尚未收到礼物的用户：

select
  b.ownerid as userid, 
  b.rebets_count, 
  b.rebets_count / 10 as gifts_expected, 
  coalesce(g.gifts_count, 0) as gifts_received,
  b.rebets_count / 10 - coalesce(g.gifts_count, 0) as gifts_missing 
from
(
  select owner_id, count(*) as rebets_count
  from bets 
  group by owner_id 
) b  
left join 
(
  select user_id, count(*) as gifts_count
  from gifts 
  where type = 'REBET_GIFT' 
  group by user_id 
) g on g.user_id = b.owner_id;

Postgres：根据结果加入两个查询和 select

Postgres: join two queries and select based on result

sql

postgresql

inner-join

subquery

数据样本：