Selecting rows ordered by some column and distinct on another

问题

Related to - PostgreSQL DISTINCT ON with different ORDER BY

I have table purchases (product_id, purchased_at, address_id)

Sample data:

| id | product_id |   purchased_at    | address_id |
| 1  |     2      | 20 Mar 2012 21:01 |     1      |
| 2  |     2      | 20 Mar 2012 21:33 |     1      |
| 3  |     2      | 20 Mar 2012 21:39 |     2      |
| 4  |     2      | 20 Mar 2012 21:48 |     2      |

The result I expect is the most recent purchased product (full row) for each address_id and that result must be sorted in descendant order by the purchased_at field:

| id | product_id |   purchased_at    | address_id |
| 4  |     2      | 20 Mar 2012 21:48 |     2      |
| 2  |     2      | 20 Mar 2012 21:33 |     1      |

Using query:

SELECT DISTINCT ON (address_id) purchases.address_id, purchases.*
FROM "purchases"
WHERE "purchases"."product_id" = 2
ORDER BY purchases.address_id ASC, purchases.purchased_at DESC

I'm getting:

| id | product_id |   purchased_at    | address_id |
| 2  |     2      | 20 Mar 2012 21:33 |     1      |
| 4  |     2      | 20 Mar 2012 21:48 |     2      |

So the rows is same, but order is wrong. Any way to fix it?

回答1:

Quite a clear question :)

SELECT t1.* FROM purchases t1
LEFT JOIN purchases t2
ON t1.address_id = t2.address_id AND t1.purchased_at < t2.purchased_at
WHERE t2.purchased_at IS NULL
ORDER BY t1.purchased_at DESC

And most likely a faster approach:

SELECT t1.* FROM purchases t1
JOIN (
    SELECT address_id, max(purchased_at) max_purchased_at
    FROM purchases
    GROUP BY address_id
) t2
ON t1.address_id = t2.address_id AND t1.purchased_at = t2.max_purchased_at
ORDER BY t1.purchased_at DESC

回答2:

Your ORDER BY is used by DISTINCT ON for picking which row for each distinct address_id to produce. If you then want to order the resulting records, make the DISTINCT ON a subselect and order its results:

SELECT * FROM
(
  SELECT DISTINCT ON (address_id) purchases.address_id, purchases.*
  FROM "purchases"
  WHERE "purchases"."product_id" = 2
  ORDER BY purchases.address_id ASC, purchases.purchased_at DESC
) distinct_addrs
order by distinct_addrs.purchased_at DESC

回答3:

This query is trickier to rephrase properly than it looks.

The currently accepted, join-based answer doesn’t correctly handle the case where two candidate rows have the same given purchased_at value: it will return both rows.

You can get the right behaviour this way:

SELECT * FROM purchases AS given
WHERE product_id = 2
AND NOT EXISTS (
    SELECT NULL FROM purchases AS other
    WHERE given.address_id = other.address_id
    AND (given.purchased_at < other.purchased_at OR given.id < other.id)
)
ORDER BY purchased_at DESC

Note how it has a fallback of comparing id values to disambiguate the case in which the purchased_at values match. This ensures that the condition can only ever be true for a single row among those that have the same address_id value.

The original query using DISTINCT ON handles this case automatically!

Also note the way that you are forced to encode the fact that you want “the latest for each address_id” twice, both in the given.purchased_at < other.purchased_at condition and the ORDER BY purchased_at DESC clause, and you have to make sure they match. I had to spend a few extra minutes to convince myself that this query is really positively correct.

It’s much easier to write this query correctly and understandbly by using DISTINCT ON together with an outer subquery, as suggested by dbenhur.

来源：https://stackoverflow.com/questions/9796078/selecting-rows-ordered-by-some-column-and-distinct-on-another

标签

postgresql

greatest-n-per-group

distinct-on