我想运行这个查询:
SELECT DISTINCT ON (address_id) purchases.address_id, purchases.*
FROM purchases
WHERE purchases.product_id = 1
ORDER BY purchases.purchased_at DESC
但是我得到了这个错误:
SELECT DISTINCT ON表达式必须匹配表达式的初始顺序
添加address_id作为第一个ORDER BY表达式可以消除错误，但我真的不想在address_id上添加排序。有没有可能不按address_id排序?
当前回答
Window函数可以一次性解决这个问题:
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)
其他回答
文档中说:
DISTINCT ON(表达式[，…])只保留给定表达式求值为相等的每一组行的第一行。[…注意，每个集合的“第一行”是不可预测的，除非使用ORDER BY来确保所需的行首先出现。[…DISTINCT ON表达式必须匹配最左边的ORDER BY表达式。
官方文档
因此，您必须将address_id添加到order中。
或者，如果你正在寻找包含每个address_id的最近购买的产品的完整行，并且结果按purchased_at排序，那么你正在尝试解决每组最大N的问题，可以通过以下方法解决:
适用于大多数dbms的通用解决方案:
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
基于@hkf的回答，一个更面向postgresql的解决方案:
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
此处澄清、扩展和解决的问题:按某列排序并在另一列上不同地选择行
你可以在子查询中根据address_id排序，然后在外部查询中根据你想要的排序。
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
你也可以通过使用group by子句来做到这一点
SELECT purchases.address_id, purchases.* FROM "purchases"
WHERE "purchases"."product_id" = 1 GROUP BY address_id,
purchases.purchased_at ORDER purchases.purchased_at DESC
它也可以使用以下查询和其他答案来解决。
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
对于任何使用Flask-SQLAlchemy的人来说，这对我来说都是有效的
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))
