3

Is there a way to set values in specific positions inside an array, based on information from other columns? (Postgres 9.3 or later.)

For example, I would like to select an item and its stock information from the following tables:

Table item:

CREATE TABLE item (
  id integer NOT NULL
);

INSERT INTO item VALUES
 (1), (2), (3), (4);

Table item_stock (containing shop-specific information like stock and prices):

CREATE TABLE item_stock (
    item_id integer NOT NULL,
    shop_id integer NOT NULL,
    stock integer,
    cost numeric(19,3),
);

INSERT INTO items_stock VALUES
  (1, 1, 2, 10),
  (1, 2, 0, 9),
  (2, 2, 0, 9),
  (3, 1, 3, 22);

SQLFiddle

Looking for a query to produce the following results, where the array in the column stock contains stock info for specific shops. In the example, array position 1 is stock for shop_id=1 and array position 2 is stock for shop_id=2. 0 instead of NULL where no data is found:

id | stock
---+-------
1  | {2, 0}
2  | {0, 0}
3  | {3, 0}
4  | {0, 0}

2 Answers 2

1

Your answer basically gets the job done:

SELECT b.id, array_agg(b.stock) AS stock
FROM  (
   SELECT i.id, COALESCE(i_s.stock, 0) AS stock
   FROM   item i
   CROSS  JOIN unnest('{1,2}'::int[]) n
   LEFT   JOIN item_stock i_s ON i.id = i_s.item_id AND n.n = i_s.shop_id
   ORDER  BY i.id, n.n
   ) b
GROUP  BY b.id;

Two notable changes:

  1. Order is not guaranteed without ORDER BY in the subquery or as additional clause to array_aggregate() (typically more expensive). And that's the core element of your question.

  2. unnest('{1,2}'::int[]) instead of generate_series(1,2) as requested shop IDs will hardly be sequential all the time.

I also moved the set-returning function from the SELECT list to a separate table expression attached with CROSS JOIN. Standard SQL form, but that's just a matter of clarity and taste, not a necessity. At least in Postgres 10 or later. See:

Doing the same with LEFT JOIN LATERAL and an ARRAY constructor might be a bit faster as we don't need the outer GROUP BY and the ARRAY constructor is typically faster, too:

SELECT i.id, s.stock
FROM   item i
CROSS  JOIN LATERAL (
   SELECT ARRAY(
      SELECT COALESCE(i_s.stock, 0)
      FROM   unnest('{1,2}'::int[]) n
      LEFT   JOIN item_stock i_s ON i_s.shop_id = n.n
                                AND i_s.item_id = i.id
      ORDER  BY n.n
      ) AS stock
   ) s;

Related:

And if you have more than just the two shops, a nested crosstab() should provide top performance:

SELECT i.id, COALESCE(stock, '{0,0}') AS stock
FROM   item i
LEFT   JOIN (
   SELECT id, ARRAY[COALESCE(shop1, 0), COALESCE(shop2, 0)] AS stock
   FROM   crosstab(
     $$SELECT item_id, shop_id, stock
       FROM   item_stock
       WHERE  shop_id = ANY ('{1,2}'::int[])
       ORDER  BY 1,2$$

     , $$SELECT unnest('{1,2}'::int[])$$
      ) AS ct (id int, shop1 int, shop2 int)
   ) i_s USING (id);

Needs to be adapted in more places to cater for different shop IDs.

Related:

db<>fiddle here

Index

Make sure you have at least an index on item_stock (shop_id, item_id) - typically provided by a PRIMARY KEY on those columns. For the crosstab query, it also matters that shop_id comes first. See:

Adding stock as another index expression may allow faster index-only scans. In Postgres 11 or later consider an INCLUDE item to the PK like so:

PRIMARY KEY (shop_id, item_id) INCLUDE (stock)

But only if you need it a lot, as it makes the index a bit bigger and possibly more susceptible to bloat from updates.

1

This is the query I was able to come up (with some brute-force):

SELECT b.id, array_agg(b.stock) FROM (
  SELECT a.*, COALESCE(i_s.stock, 0) as stock FROM (
    SELECT id, generate_series(1, 2) as n FROM items  
  ) as a
  LEFT OUTER JOIN item_stock i_s ON a.id = i_s.item_id AND a.n = i_s.shop_id
) as b GROUP by b.id;

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.