PostgreSQL DISTINCT ON与不同的ORDER BY

Question

我想运行此查询:

SELECT DISTINCT ON (address_id) purchases.address_id, purchases.*
FROM purchases
WHERE purchases.product_id = 1
ORDER BY purchases.purchased_at DESC

但我得到这个错误:

PG ::错误:错误:SELECT DISTINCT ON表达式必须与初始ORDER BY表达式匹配

添加address_id为第一个ORDER BY表达式会使错误无效,但我真的不想添加排序address_id.是否可以不通过订购address_id？

Answer 1

文件说:

DISTINCT ON(expression [,...])仅保留给定表达式求值的每组行的第一行.[...]请注意,每个集合的"第一行"是不可预测的,除非使用ORDER BY来确保首先出现所需的行.[...] DISTINCT ON表达式必须与最左边的ORDER BY表达式匹配.

官方文件

因此,您必须将address_id订单添加到订单中.

或者,如果您正在寻找包含最新购买的产品的完整行,address_id并且该结果按purchased_at那时排序,则您尝试解决每组最大的N问题,可以通过以下方法解决:

应该适用于大多数DBMS的一般解决方案:

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

一个更基于PostgreSQL的解决方案基于@hkf的答案:

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

问题在此处得到澄清,扩展和解决:选择由某些列排序的行和在另一列上排序的行

Answer 2

您可以通过子查询中的address_id进行排序,然后按外部查询中的内容进行排序.

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

Answer 3

一个子查询可以解决这个问题:

SELECT *
FROM  (
    SELECT DISTINCT ON (address_id) *
    FROM   purchases
    WHERE  product_id = 1
    ) p
ORDER  BY purchased_at DESC;

领先的表达式ORDER BY必须与列中的列一致DISTINCT ON,因此您不能在同一列中按不同的顺序排序SELECT.

ORDER BY如果要从每个集合中选择特定行,则仅在子查询中使用其他行:

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;

如果purchased_at可以NULL,请考虑DESC NULLS LAST.
相关,有更多解释:

• 选择每个GROUP BY组中的第一行？
• PostgreSQL按日期时间asc排序,先是null吗？

Answer 4

窗函数可以一次解决:

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)

Answer 5

对于使用Flask-SQLAlchemy 的任何人，这对我有用

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))