ruby-on-rails - Rails 每天只保存众多记录中的 1 条。保留最后，删除其余

Question

Rails 3.1、Ruby 1.9.2、AR/MySQL。

我正在寻找有关如何在每个时间段(天)内仅保留 1 个结果的建议，如果同一类型的结果在该时间段内有很多结果。一个例子可能是跟踪股票价格。最初，我们每 15 分钟保存一次价格，但只需要将每个价格点存储 1 周。第一周后，我们每天只需要一个价格(最后记录，收盘价)。

这是一个简单的第一次尝试，它确实有效，但效率极低:

# stock has many prices, price has one stock
# get all prices for single stock older than 1 week
prices = stock.prices.where("created_at < ? ", Time.now-1.week)  
prices.group_by{ |price| price.created_at.to_date }.each do |k,v| # group by day
  if v.count > 1  # if many price points that day
    (v[0]..v[v.size-2]).each {|r| r.delete} # delete all but last record in day
  end
end

提前感谢您提供的任何帮助/建议。我会在完成过程中尝试更新，希望它能对以后的人有所帮助。

Answer 1

您可以通过在 SQL 中完成所有操作并将范围限制为上次运行的时间来提高效率。此外，如果您添加一列以将较旧的日终条目标记为“已存档”，那么它会使查询变得更加简单。存档价格是您不会在一周后删除的价格。

rails generate migration add_archived_to_prices archived:boolean

在迁移之前，将迁移修改为在 created_at 列上建立索引。

class AddArchivedToPrices < ActiveRecord::Migration
  def self.up
    add_column :prices, :archived, :boolean
    add_index :prices, :created_at
  end

  def self.down
    remove_index :prices, :created_at
    remove_column :prices, :archived
  end
end

工作流程是这样的:

# Find the last entry for each day for each stock using SQL (more efficient than finding these in Ruby)
keepers =
  Price.group('stock_id, DATE(created_at)').
        having('created_at = MAX(created_at)').
        select(:id).
        where('created_at > ?', last_run) # Keep track of the last run time to speed up subsequent runs

# Mark them as archived
Price.where('id IN (?)', keepers.map(&:id)).update_all(:archived => true)

# Delete everything but archived prices that are older than a week
Price.where('archived != ?', true).
      where('created_at < ?", Time.now - 1.week).
      where('created_at > ?', last_run). # Keep track of the last run time to speed up subsequent runs
      delete_all

最后要注意的是，一定不要将 group() 和 update_all() 组合在一起。 group() 被 update_all() 忽略。