Postgresql – Getting the first 3 children(items) for each owner in a list of primary keys

Something like this:

select tc.table_schema, tc.table_name, kc.column_name
from information_schema.table_constraints tc
  join information_schema.key_column_usage kc 
    on kc.table_name = tc.table_name and kc.table_schema = tc.table_schema and kc.constraint_name = tc.constraint_name
where tc.constraint_type = 'PRIMARY KEY'
  and kc.ordinal_position is not null
order by tc.table_schema,
         tc.table_name,
         kc.position_in_unique_constraint;

Consistent rows

The important question which does not seem to be on your radar yet:
From each set of rows for the same seriesName, do you want the columns of one row, or just any values from multiple rows (which may or may not go together)?

Your answer does the latter, you combine the maximum dbid with the maximum retreivaltime, which may come from a different row.

To get consistent rows, use DISTINCT ON and wrap it in a subquery to order the result differently:

SELECT * FROM (
   SELECT DISTINCT ON (seriesName)
          dbid, seriesName, retreivaltime
   FROM   FileItems
   WHERE  sourceSite = 'mk' 
   ORDER  BY seriesName, retreivaltime DESC NULLS LAST  -- latest retreivaltime
   ) sub
ORDER BY retreivaltime DESC NULLS LAST
LIMIT  100;

Details for DISTINCT ON:

• Select first row in each GROUP BY group?

Aside: should probably be retrievalTime, or better yet: retrieval_time. Unquoted mixed case identifiers are a common source of confusion in Postgres.

Better Performance with rCTE

Since we are dealing with a big table here, we'd need a query that can use an index, which is not the case for the above query (except for WHERE sourceSite = 'mk')

On closer inspection, your problem seems to be a special case of a loose index scan. Postgres does not support loose index scans natively, but it can be emulated with a recursive CTE. There is a code example for the simple case in the Postgres Wiki.

Related answer on SO with more advanced solutions, explanation, fiddle:

• Optimize GROUP BY query to retrieve latest record per user

Your case is more complex, though. But I think I found a variant to make it work for you. Building on this index (without WHERE sourceSite = 'mk')

CREATE INDEX mi_special_full_idx ON MangaItems
(retreivaltime DESC NULLS LAST, seriesName DESC NULLS LAST, dbid)

Or (with WHERE sourceSite = 'mk')

CREATE INDEX mi_special_granulated_idx ON MangaItems
(sourceSite, retreivaltime DESC NULLS LAST, seriesName DESC NULLS LAST, dbid)

The first index can be used for both queries, but is not fully efficient with the additional WHERE condition. The second index is of very limited use for the first query. Since you have both variants of the query consider creating both indexes.

I added dbid at the end to allow Index Only scans.

This query with a recursive CTE makes use of the index. I tested with Postgres 9.3 and it works for me: no sequential scan, all index-only scans:

WITH RECURSIVE cte AS (
   (
   SELECT dbid, seriesName, retreivaltime, 1 AS rn, ARRAY[seriesName] AS arr
   FROM   MangaItems
   WHERE  sourceSite = 'mk'
   ORDER  BY retreivaltime DESC NULLS LAST, seriesName DESC NULLS LAST
   LIMIT  1
   )
   UNION ALL
   SELECT i.dbid, i.seriesName, i.retreivaltime, c.rn + 1, c.arr || i.seriesName
   FROM   cte c
   ,      LATERAL (
      SELECT dbid, seriesName, retreivaltime
      FROM   MangaItems
      WHERE (retreivaltime, seriesName) < (c.retreivaltime, c.seriesName)
      AND    sourceSite = 'mk'  -- repeat condition!
      AND    seriesName <> ALL(c.arr)
      ORDER  BY retreivaltime DESC NULLS LAST, seriesName DESC NULLS LAST
      LIMIT  1
      ) i
   WHERE  c.rn < 101
   )
SELECT dbid
FROM   cte
ORDER  BY rn;

You need to include seriesName in ORDER BY, since retreivaltime is not unique. "Almost" unique is still not unique.

Explain

• The non-recursive query starts with the latest row.

• The recursive query adds the next-latest row with a seriesName that's not in the list, yet etc., until we have 100 rows.

• Essential parts are the JOIN condition (b.retreivaltime, b.seriesName) < (c.retreivaltime, c.seriesName) and the ORDER BY clause ORDER BY retreivaltime DESC NULLS LAST, seriesName DESC NULLS LAST. Both match the sort order of the index, which allows for the magic to happen.

• Collecting seriesName in an array to rule out duplicates. The cost for b.seriesName <> ALL(c.foo_arr) grows progressively with the number of rows, but for just 100 rows it is still cheap.

• Just returning dbid as clarified in the comments.

Alternative with partial indexes:

We have been dealing with similar problems before. Here is a highly optimized complete solution based on partial indexes and a looping function:

• Can spatial index help a "range - order by - limit" query

Probably the fastest way (except for a materialized view) if done right. But more complex.

Materialized View

Since you do not have a lot of write operations and they are not performance-critical as stated in the comments (should be in the question), save the top n pre-computed rows in a materialized view and refresh it after relevant changes to the underlying table. Base your performance-critical queries on the materialized view instead.

• Could just be a "thin" mv of the latest 1000 dbid or so. In the query, join to the original table. For instance, if content is sometimes updated, but the top n rows can remain unchanged.

• Or a "fat" mv with whole rows to return. Faster, yet. Needs to be refreshed more often, obviously.

Details in the manual here and here.