aggregation subquery with top N

Question

Let's say I have a table called COFFEE showing Bus Stations and all of the coffee shops within 10 blocks of the bus station:

    BusStationID| CoffeeShopID |  Distance (in city blocks)
    1|2103|2
    1|2222|2
    1|8864|7
    1|9920|5
    1|3544|2
    1|4830|2
    1|4823|6
    1|9561|2
    7|6262|2
    7|8561|10
    7|9510|5
    7|2744|1       
    7|4223|9
    7|5960|3

[EDITED: to make clear that the question is how to do this with a query not procedurally]

And I have to write a query (not a proc) to show for each bus-station, the average distance to the five closest coffee shops.

I can get the top 5 closest coffee shops for a particular bus-station:

           select avg(top5.distance) as AvgDistToFiveClosest
           from
           (
           select top 5 distance from COFFEE where busstationid = 1
           order by distance
           ) as top5

But how do I connect that as a subquery and make AvgDistToFiveClosest a column returned in my main query:

        select BusStationId,  AvgDistToFiveClosest
        from COFFEE...
         ??????

Given the sample data above, the query should return:

     BusStationID | AvgDistToFiveClosest
           1 | 2
           7 | 4

Answer 1

Try this:

SELECT c.BusStationID, AVG(c.distance)
FROM COFFEE c
WHERE c.CoffeeShopID IN 
(SELECT TOP 5 c2.CoffeeShopID FROM COFFEE c2 WHERE c2.BusStationID = c.BusStationID
ORDER BY c2.distance)
GROUP BY c.BusStationID

Answer 2

This is Oracle (9g+) SQL code, corrected, I found an answer for a single select statement

with
distanceRanks as
(
  SELECT
    busstationid,
    distance,
    --rank the rows on the distance column from smallest to longest, and differentiate equal distances by rownum
    rank() over ( partition by busstationid
                  order by distance, rownum asc) as ranking
  FROM coffee
  ORDER BY 1 asc
)
SELECT busstationid, avg(distance)
FROM distanceRanks
WHERE ranking < 6
group by busstationid;

Answer 3

Try this

Select distinct busstationid , (select top 5 avg(distance) from coffee ce where ce.busstationid = b.busstationid order by distance) as AvgDistToFiveClosest
From coffee b