PostgreSQL 具有不同 ORDER BY 的 DISTINCT ON

Question

我想運行這個查詢：

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

但我收到此錯誤：

PG::Error: ERROR: SELECT DISTINCT ON 表達式必須匹配初始 ORDER BY 表達式

添加address_id作為第一個ORDER BY表達式可以消除錯誤，但我真的不想在address_id上添加排序。 是否可以不通過address_id訂購？

Answer 1

文檔說：

DISTINCT ON ( expression [, ...] ) 僅保留給定表達式計算結果相等的每組行的第一行。 [...] 請注意，每個集合的“第一行”是不可預測的，除非使用 ORDER BY 來確保所需的行首先出現。 [...] DISTINCT ON 表達式必須匹配最左邊的 ORDER BY 表達式。

官方文檔

因此，您必須將address_id添加到 order by 中。

或者，如果您正在尋找包含每個address_id最新購買產品的完整行，並且該結果按purchased_at排序，那么您正在嘗試解決每組最大 N 問題，該問題可以通過以下方法解決：

適用於大多數 DBMS 的通用解決方案：

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

基於@hkf 的答案的更面向 PostgreSQL 的解決方案：

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

問題在此處得到澄清、擴展和解決：選擇按某列排序並在另一列上不同的行

Answer 2

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

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

子查詢可以解決它：

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

ORDER BY前導表達式必須與DISTINCT ON列一致，因此您不能在同一個SELECT按不同列進行排序。

如果要從每個集合中選擇特定行，請僅在子查詢中使用額外的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 ，請使用DESC NULLS LAST - 並匹配您的索引以獲得最佳性能。 看：

• 按列 ASC 排序，但首先是 NULL 值？
• 為什么 ORDER BY NULLS LAST 會影響主鍵上的查詢計划？

相關，有更多解釋：

• 選擇每個 GROUP BY 組中的第一行？
• 按列 ASC 排序，但首先是 NULL 值？

Answer 4

窗口函數可以一次性解決這個問題：

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

對於使用Flask-SQLAlchemy 的任何人，這對我有用

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

對於DISTINCT ON（）函數，必須按順序添加address_id

Answer 7

也可以使用以下查詢和其他答案來解決。

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

試試這個查詢。

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 子句來完成此操作

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