PostgreSQL 具有不同 ORDER BY 的 DISTINCT ON

Question

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

但我收到此错误：

PG::Error: ERROR: SELECT DISTINCT ON 表达式必须匹配初始 ORDER BY 表达式

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。 是否可以不通过address_id订购？

Answer 1

文档说：

DISTINCT ON ( expression [, ...] ) 仅保留给定表达式计算结果相等的每组行的第一行。 [...] 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 by 中。

或者，如果您正在寻找包含每个address_id最新购买产品的完整行，并且该结果按purchased_at排序，那么您正在尝试解决每组最大 N 问题，该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 的答案的更面向 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

问题在此处得到澄清、扩展和解决：选择按某列排序并在另一列上不同的行

Answer 2

您可以在子查询中按 address_id 排序，然后在外部查询中按您想要的排序。

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

Answer 3

子查询可以解决它：

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

ORDER BY前导表达式必须与DISTINCT ON列一致，因此您不能在同一个SELECT按不同列进行排序。

如果要从每个集合中选择特定行，请仅在子查询中使用额外的ORDER BY ：

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可以是NULL ，请使用DESC NULLS LAST - 并匹配您的索引以获得最佳性能。 看：

• 按列 ASC 排序，但首先是 NULL 值？
• 为什么 ORDER BY NULLS LAST 会影响主键上的查询计划？

相关，有更多解释：

• 选择每个 GROUP BY 组中的第一行？
• 按列 ASC 排序，但首先是 NULL 值？

Answer 4

窗口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)

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))

Answer 6

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

ORDER BY address_id ，purchases.purchased_at DESC

对于DISTINCT ON（）函数，必须按顺序添加address_id

Answer 7

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

Answer 8

试试这个查询。

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

Answer 9

您也可以使用 group by 子句来完成此操作

   SELECT purchases.address_id, purchases.* FROM "purchases"
    WHERE "purchases"."product_id" = 1 GROUP BY address_id,
purchases.purchased_at ORDER purchases.purchased_at DESC