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

添加address_id为第一ORDER BY表达沉默的错误，但我真的不想在address_id上添加排序。没有address_id订购可以吗？

Answer 1

文件说：

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

所以你必须给address_id添加到由顺序。

或者，如果您正在查找包含每个address_id的最新购买产品的整行，并且结果按purchased_at排序，那么您正试图解决最大的每组问题，可以通过以下方法：

一般的解决方案，应该在大多数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的解决方案基于香港小轮@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 

问题澄清，扩展并在这里解决：Selecting rows ordered by some column and distinct on another

Answer 2

您可以通过address_id在子查询中进行排序，然后按照您希望在外部查询中进行排序。

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; 

在ORDER BY领先的词句在DISTINCT ON与列同意，所以不能按订单不同的列在相同的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，考虑DESC NULLS LAST：

只有在子查询，如果你想从每组选择一个特定的行使用附加ORDER BY。
相关，与更多的解释：

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

Answer 4

窗口功能可以解决一通：

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

对于使用烧瓶SQLAlchemy的人，这个工作对我来说

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

您也可以通过使用GROUP BY子句这样做

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