PostgreSQL的DISTINCT與不同的ORDER BY

Question

我想運行此查詢：

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

但我得到這個錯誤：

PG::Error: ERROR: SELECT DISTINCT ON expressions must match initial ORDER BY expressions

添加address_id爲第一ORDER BY表達沉默的錯誤，但我真的不想在address_id上添加排序。沒有address_id訂購可以嗎？

Answer 1

文件說：

DISTINCT ON (expression [, ...]) keeps only the first row of each set of rows where the given expressions evaluate to equal. [...] Note that the "first row" of each set is unpredictable unless ORDER BY is used to ensure that the desired row appears first. [...] The DISTINCT ON expression(s) must match the leftmost ORDER BY expression(s).

Official documentation

所以你必須給address_id添加到由順序。

或者，如果您正在查找包含每個address_id的最新購買產品的整行，並且結果按purchased_at排序，那麼您正試圖解決最大的每組問題，可以通過以下方法：

一般的解決方案，應該在大多數DBMS的工作：

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 

更加面向PostgreSQL的解決方案基於香港小輪@的回答：

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 

問題澄清，擴展並在這裏解決：Selecting rows ordered by some column and distinct on another

Answer 2

您可以通過address_id在子查詢中進行排序，然後按照您希望在外部查詢中進行排序。

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

一個子查詢可以解決這個問題：

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

在ORDER BY領先的詞句在DISTINCT ON與列同意，所以不能按訂單不同的列在相同的SELECT。

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，考慮DESC NULLS LAST：

只有在子查詢，如果你想從每組選擇一個特定的行使用附加ORDER BY。
相關，與更多的解釋：

• Select first row in each GROUP BY group?
• PostgreSQL sort by datetime asc, null first?

Answer 4

窗口功能可以解決一通：

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

對於使用燒瓶SQLAlchemy的人，這個工作對我來說

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

您也可以通過使用GROUP BY子句這樣做

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