Documentation says:
DISTINCT ON ( expression [, ...] ) keeps only the first row of each set of rows where the given expressions evaluate to equal. [...] Note that the "first row" of each set is unpredictable unless ORDER BY is used to ensure that the desired row appears first. [...] The DISTINCT ON expression(s) must match the leftmost ORDER BY expression(s).
Official documentation
So you'll have to add the address_id
to the order by.
Alternatively, if you're looking for the full row that contains the most recent purchased product for each address_id
and that result sorted by purchased_at
then you're trying to solve a greatest N per group problem which can be solved by the following approaches:
The general solution that should work in most DBMSs:
SELECT t1.* FROM purchases t1
JOIN (
SELECT address_id, max(purchased_at) max_purchased_at
FROM purchases
WHERE product_id = 1
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
A more PostgreSQL-oriented solution based on @hkf's answer:
SELECT * FROM (
SELECT DISTINCT ON (address_id) *
FROM purchases
WHERE product_id = 1
ORDER BY address_id, purchased_at DESC
) t
ORDER BY purchased_at DESC
Problem clarified, extended and solved here: Selecting rows ordered by some column and distinct on another
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