Select unique values sorted by date - postgresql

Question

I am trying to solve an interesting problem. I have a table that has, among other data, these columns (dates in this sample are shown in European format - dd/mm/yyyy):
n_place_id dt_visit_date
(integer) (date)
========== =============
1 10/02/2012
3 11/03/2012
4 11/05/2012
13 14/06/2012
3 04/10/2012
3 03/11/2012
5 05/09/2012
13 18/08/2012
Basically, each place may be visited multiple times - and the dates may be in the past (completed visits) or in the future (planned visits). For the sake of simplicity, today's visits are part of planned future visits.
Now, I need to run a select on this table, which would pull unique place IDs from this table (without date) sorted in the following order:
Future visits go before past visits
Future visits take precedence in sorting over past visits for the same place
For future visits, the earliest date must take precedence in sorting for the same place
For past visits, the latest date must take precedence in sorting for the same place.
For example, for the sample data shown above, the result I need is:
5 (earliest future visit)
3 (next future visit into the future)
13 (latest past visit)
4 (previous past visit)
1 (earlier visit in the past)
Now, I can achieve the desired sorting using case when in the order by clause like so:
select
n_place_id
from
place_visit
order by
(case when dt_visit_date >= now()::date then 1 else 2 end),
(case when dt_visit_date >= now():: date then 1 else -1 end) * extract(epoch from dt_visit_date)
This sort of does what I need, but it does contain repeated IDs, whereas I need unique place IDs. If I try to add distinct to the select statement, postgres complains that I must have the order by in the select clause - but then the unique won't be sensible any more, as I have dates in there.
Somehow I feel that there should be a way to get the result I need in one select statement, but I can't get my head around how to do it.
If this can't be done, then, of course, I'll have to do the whole thing in the code, but I'd prefer to have this in one SQL statement.
P.S. I am not worried about the performance, because the dataset I will be sorting is not large. After the where clause will be applied, it will rarely contain more than about 10 records.

Answer 1

With DISTINCT ON you can easily show additional columns of the row with the resulting n_place_id:
SELECT n_place_id, dt_visit_date
FROM (
SELECT DISTINCT ON (n_place_id) *
,dt_visit_date < now()::date AS prio -- future first
,#(now()::date - dt_visit_date) AS diff -- closest first
FROM place_visit
ORDER BY n_place_id, prio, diff
) x
ORDER BY prio, diff;
Effectively I pick the row with the earliest future date (including "today") per n_place_id - or latest date in the past, failing that.
Then the resulting unique rows are sorted by the same criteria.
FALSE sorts before TRUE
The "absolute value" # helps to sort "closest first"
More on the Postgres specific DISTINCT ON in this related answer.
Result:
n_place_id | dt_visit_date
------------+--------------
5 | 2012-09-05
3 | 2012-10-04
13 | 2012-08-18
4 | 2012-05-11
1 | 2012-02-10

Answer 2

Try this
select n_place_id
from
(
select *,
extract(epoch from (dt_visit_date - now())) as seconds,
1 - SIGN(extract(epoch from (dt_visit_date - now())) ) as futurepast
from #t
) v
group by n_place_id
order by max(futurepast) desc, min(abs(seconds))