Question

SELECT distinct on (prices.item_id) *
FROM prices
ORDER BY prices.item_id, prices.updated_at DESC

以上查询检索最近的价格，我如何得到所有当前价格的总和？

是否可以不使用子选择？

Answer 1

使用子查询这是微不足道的：

select sum(p.price)
from (select distinct on (p.item_id) p.*
      from prices p
      order by p.item_id, p.updated_at desc
     ) p

如果您不介意重复行，我认为以下内容可能会有效：

select distinct on (p.item_id) sum(prices.price) over ()
from prices p
order by p.item_id, p.updated_at desc

您可以为此添加限制条款以获得所需内容。顺便说一下，我会把它写成：

select sum(p.price)
from (select p.*,
             row_number() over (partition by p.item_id order by updated_at desc) as seqnum
      from prices p
      order by p.item_id, p.updated_at desc
     ) p
where seqnum = 1

ROW_NUMBER（）是标准SQL。 DISTINCT ON子句特定于Postgres。

汇总最新的行，按ID分组

1 个答案: