SELECT DISTINCT ON (address_id) purchases.address_id, purchases.*
FROM purchases
WHERE purchases.product_id = 1
ORDER BY purchases.purchased_at DESC
But I get this error:
PG::Error: ERROR: SELECT DISTINCT ON expressions must match initial ORDER BY expressions
Adding address_id as first ORDER BY expression silences the error, but I really don't want to add sorting over address_id. Is it possible to do without ordering by address_id?
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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
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
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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ORDER BY purchased_at DESC;
Leading expressions in ORDER BY have to agree with columns in DISTINCT ON, so you can't order by different columns in the same SELECT.
Only use an additional ORDER BY in the subquery if you want to pick a particular row from each set:
SELECT *
FROM (
SELECT DISTINCT ON (address_id) *
FROM purchases
WHERE product_id = 1
ORDER BY address_id, purchased_at DESC -- get "latest" row per address_id
ORDER BY purchased_at DESC;
If purchased_at can be NULL, use DESC NULLS LAST - and match your index for best performance. See:
Sort by column ASC, but NULL values first?
Why does ORDER BY NULLS LAST affect the query plan on a primary key?
Related, with more explanation:
Select first row in each GROUP BY group?
Sort by column ASC, but NULL values first?
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You can order by address_id in an subquery, then order by what you want in an outer query.
SELECT * FROM
(SELECT DISTINCT ON (address_id) purchases.address_id, purchases.*
FROM "purchases"
WHERE "purchases"."product_id" = 1 ORDER BY address_id DESC )
ORDER BY purchased_at DESC
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SELECT DISTINCT ON (address_id)
LAST_VALUE(purchases.address_id) OVER wnd AS address_id
FROM "purchases"
WHERE "purchases"."product_id" = 1
WINDOW wnd AS (
PARTITION BY address_id ORDER BY purchases.purchased_at DESC
ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING)
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For anyone using Flask-SQLAlchemy, this worked for me
from app import db
from app.models import Purchases
from sqlalchemy.orm import aliased
from sqlalchemy import desc
stmt = Purchases.query.distinct(Purchases.address_id).subquery('purchases')
alias = aliased(Purchases, stmt)
distinct = db.session.query(alias)
distinct.order_by(desc(alias.purchased_at))
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It can also be solved using the following query along with other answers.
WITH purchase_data AS (
SELECT address_id, purchased_at, product_id,
row_number() OVER (PARTITION BY address_id ORDER BY purchased_at DESC) AS row_number
FROM purchases
WHERE product_id = 1)
SELECT address_id, purchased_at, product_id
FROM purchase_data where row_number = 1
You can also done this by using group by clause
SELECT purchases.address_id, purchases.* FROM "purchases"
WHERE "purchases"."product_id" = 1 GROUP BY address_id,
purchases.purchased_at ORDER purchases.purchased_at DESC
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