How to rank users and get a subset from this rank with my user and the above and below user by rank position - sql

Question

I am working on a query right now to get a ranking of my users. I have two tables one for users and the other one for profits where I save the amount and the user id to which is related. By getting the total of profits generated by a user I need to build a rank with three users, the user in the next higher ranked position to my user, my user and the user in the next lower ranked position to my user. For example:
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
In this if my user is Kobe Bryant I would need to get a rank with Michael Jordan, Kobe Bryant and Phil Jackson.
If my user is Mike Bibby I would need to get a rank with Michale Jordan, Mike Bybby and Phil Jackson.
Until now I have a query that returns me a full rank with all the users but I do not now what is a good way to get the three users that I want. I have tried to do this with ruby but I think it would be better if I do all this processing in the DB.
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;
I am using 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
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