I want to run this query:
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
?
This question is related to
sql
postgresql
sql-order-by
distinct-on
A subquery can solve it:
SELECT *
FROM (
SELECT DISTINCT ON (address_id) *
FROM purchases
WHERE product_id = 1
) p
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
) p
ORDER BY purchased_at DESC;
If purchased_at
can be NULL
, use DESC NULLS LAST
- and match your index for best performance. See:
Related, with more explanation:
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 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
Window function may solve that in one pass:
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)
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
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))
Source: Stackoverflow.com