Question

我正在处理一个查询，以获取用户的排名。我有两个表用于用户，另一个用于利润，我保存金额和与之相关的用户ID。通过获得用户生成的利润总额，我需要建立一个具有三个用户的排名，用户在下一个更高排名的位置，我的用户，我的用户和用户在下一个排名较低的位置给我的用户。例如：

  id   |            name             | total_profit | rank
-------+-----------------------------+--------------+------
 10312 | John Doe                    |       7000.0 |    1
 10329 | Michael Jordan              |       5000.0 |    2
 10333 | Kobe Bryant                 |       4000.0 |    3
 10327 | Mike Bibby                  |       4000.0 |    3
 10331 | Phil Jackson                |       1000.0 |    4

如果我的用户是Kobe Bryant，我需要得到Michael Jordan，Kobe Bryant和Phil Jackson的排名。

如果我的用户是Mike Bibby，我需要获得Michale Jordan，Mike Bybby和Phil Jackson的排名。

到目前为止，我有一个查询，它让所有用户都获得了满级排名，但我现在不知道什么是获得我想要的三个用户的好方法。我试图用ruby做这个，但我认为如果我在数据库中完成所有这些处理会更好。

SELECT users.id, users.name, total_profit, rank() OVER(ORDER BY total_profit DESC)
FROM users
INNER JOIN (SELECT sum(profits.amount) AS total_profit, investor_id
          FROM profits GROUP BY profits.investor_id) profits ON profits.investor_id = users.id
ORDER BY total_profit DESC;

我正在使用PostgresSQL 9.1.4

Answer 1

with s as (
    select
        users.id, users.name, total_profit,
        rank() over(order by total_profit desc) as r
    from
        users
        inner join
            (
                select sum(profits.amount) as total_profit,
                investor_id
                from profits
                group by profits.investor_id
            ) profits on profits.investor_id = users.id
), u as (
    select r from s where name = 'Kobe Bryant'
)
select distinct on (r) id, name, total_profit, r
from s
where
    name = 'Kobe Bryant'
    or r in (
        (select r from u) - 1, (select r from u) + 1
    )
order by r;

Answer 2

with cte_profits as (
    select
        sum(p.amount) as total_profit, p.investor_id
    from profits as p
    group by p.investor_id
), cte_users_profits as (
    select
        u.id, u.name, p.toral_profit,
        dense_rank() over(order by p.total_profit desc) as rnk,
        row_number() over(partition by up.total_profit order by up.id) as rn
    from users as u
        inner join cte_profits as p on p.investor_id = u.id
)
select c2.*
from cte as c
   left outer join cte as c2 on
       c2.id = c.id or
       c2.rnk = c.rnk + 1 and c2.rn = 1 or
       c2.rnk = c.rnk - 1 and c2.rn = 1
where c.name = 'Kobe Bryant'
order by c2.rnk

sql fiddle demo

如何对用户进行排名并从我的用户以及排名位置的上下用户获得此排名的子集

2 个答案: