我怎么能找到一个相关的表的最古老的记录过滤通过它的一个属性？

你可以使用一种 横向子查询 其可以被描述为一个foreach循环在SQL。 他们是一个极高效的方式来选择列从一个单一的相关记录或甚至聚集体，从一群有关的记录。

为每个行订阅的数据库将会选择一个单一的排版本的命令authorized_at:

SELECT "subscriptions".*,
       "latest_version"."authorized_at" AS current_version,
       "latest_version"."id" AS current_version_id -- could be very useful
FROM   "subscriptions" 
LATERAL
  (
     SELECT   "versions"."authorized_at", "versions"."id"
     FROM     "versions"
     WHERE    "versions"."subscription_id" = "subscriptions"."id" -- lateral reference
     AND      "versions"."plan_id" = "subscriptions"."plan_id"
     AND      "versions"."status" = 'processed'
     ORDER BY "versions"."authorized_at" ASC
     LIMIT 1
  ) latest_version ON TRUE

建立横向联接在Email可以与SQL串或阿雷尔:

class Subscription < ApplicationRecord
  # Performs a lateral join and selects the 
  # authorized_at of the latest version 
  def self.with_current_version
    lateral = Version.arel_table.then do |v|
      v.project(
        v[:authorized_at],
        v[:id] # optional
      ).where(
        v[:subscription_id].eq(arel_table[:id])
          .and(v[:plan_id].eq(arel_table[:plan_id]) )
          .and(v[:status].eq('processed'))
      )
      .order(v[:authorized_at].asc)
      .take(1) # limit 1
      .lateral('latest_version ON TRUE')
    end
    lv = Arel::Table.new(:latest_version) # just a table alias
    select(
      *where(nil).arel.projections, # selects everything previously selected
      lv[:authorized_at].as("current_version"),
      lv[:id].as("current_version_id") # optional
    ).joins(lateral.to_sql)
  end
end

如果你只是想要选择的 id 和 current_version 你应该考虑采用替代选择的数据库模式，不适当地水合。

你可以使用 DISTINCT ON 过滤出行，并保持一个单一的每个订阅--第一个，每组根据的 ORDER BY 条款。

例如：

select distinct on (s.id) s.id, v.authorized_at
from subscription s
join versions v on v.subscription_id = s.id and v.plan_id = s.plan_id
where v.status = 'processed'
order by s.id, v.authorized_at

以下代码会给你的 versions 哪儿的版本是 plan_id 等于订阅的 plan_id.

@versions = Version.joins("LEFT JOIN subscriptions ON subscriptions.plan_id = versions.plan_id")

过滤记录的版本是 status

@versions = Version.joins("LEFT JOIN subscriptions ON subscriptions.plan_id = versions.plan_id").where(status: "processed")

过滤记录的版本是 status 和秩序， authorized_at 升序排列。

@versions = Version.joins("LEFT JOIN subscriptions ON subscriptions.plan_id = versions.plan_id").where(status: "processed").order(:authorized_at)

过滤记录的版本是 status 和秩序， authorized_at 在递减的顺序。

@versions = Version.joins("LEFT JOIN subscriptions ON subscriptions.plan_id = versions.plan_id").where(status: "processed").order(authorized_at: :desc)

希望这对你的作品！