如果没有找到记录，如何获取过去 7 天的记录并填充为 0？答案

【问题标题】：How to fetch records from last 7 days with padding to 0 if no records found?如果没有找到记录，如何获取过去 7 天的记录并填充为 0？
【发布时间】：2016-03-27 16:36:25
【问题描述】：

我正在尝试从过去 7 天（可能会更改以支持月份和年份）中获取记录以填充图表，并且我需要将未找到记录的日期填充为 0。我首先尝试分组但发现无法填充数据，所以我想出了这个：

today = Date.today

array_data = (today - 6.days..today).map do |day|  
  total_time = @project.time_entries.where(created_at: (day.beginning_of_day..day.end_of_day)).map(&:duration_hours).sum

  { day.strftime("%a %d %b") => total_time }
end

@entries_data = array_data.reduce Hash.new, :merge

结果是一个散列，其中一天为键，计算属性的总和为值。

但是，这需要每天进行一次数据库查询：

TimeEntry Load (1.1ms)  SELECT "time_entries".* FROM "time_entries" WHERE "time_entries"."project_id" IN ('bd9f541c-e37a-45de-bc94-28bef2b5eade')
  TimeEntry Load (0.4ms)  SELECT "time_entries".* FROM "time_entries" WHERE "time_entries"."project_id" = $1 AND ("time_entries"."created_at" BETWEEN '2016-03-21 04:30:00.000000' AND '2016-03-22 04:29:59.999999')  [["project_id", "bd9f541c-e37a-45de-bc94-28bef2b5eade"]]
  TimeEntry Load (0.2ms)  SELECT "time_entries".* FROM "time_entries" WHERE "time_entries"."project_id" = $1 AND ("time_entries"."created_at" BETWEEN '2016-03-22 04:30:00.000000' AND '2016-03-23 04:29:59.999999')  [["project_id", "bd9f541c-e37a-45de-bc94-28bef2b5eade"]]
  TimeEntry Load (0.2ms)  SELECT "time_entries".* FROM "time_entries" WHERE "time_entries"."project_id" = $1 AND ("time_entries"."created_at" BETWEEN '2016-03-23 04:30:00.000000' AND '2016-03-24 04:29:59.999999')  [["project_id", "bd9f541c-e37a-45de-bc94-28bef2b5eade"]]
  TimeEntry Load (0.2ms)  SELECT "time_entries".* FROM "time_entries" WHERE "time_entries"."project_id" = $1 AND ("time_entries"."created_at" BETWEEN '2016-03-24 04:30:00.000000' AND '2016-03-25 04:29:59.999999')  [["project_id", "bd9f541c-e37a-45de-bc94-28bef2b5eade"]]
  TimeEntry Load (0.5ms)  SELECT "time_entries".* FROM "time_entries" WHERE "time_entries"."project_id" = $1 AND ("time_entries"."created_at" BETWEEN '2016-03-25 04:30:00.000000' AND '2016-03-26 04:29:59.999999')  [["project_id", "bd9f541c-e37a-45de-bc94-28bef2b5eade"]]
  TimeEntry Load (0.6ms)  SELECT "time_entries".* FROM "time_entries" WHERE "time_entries"."project_id" = $1 AND ("time_entries"."created_at" BETWEEN '2016-03-26 04:30:00.000000' AND '2016-03-27 04:29:59.999999')  [["project_id", "bd9f541c-e37a-45de-bc94-28bef2b5eade"]]
  TimeEntry Load (0.5ms)  SELECT "time_entries".* FROM "time_entries" WHERE "time_entries"."project_id" = $1 AND ("time_entries"."created_at" BETWEEN '2016-03-27 04:30:00.000000' AND '2016-03-28 04:29:59.999999')  [["project_id", "bd9f541c-e37a-45de-bc94-28bef2b5eade"]]

我怎样才能修改它以使其仅向数据库发送 1 个查询？

【问题讨论】：

标签： ruby-on-rails activerecord

【解决方案1】：

不确定这是您的意思，但我认为您需要这样做：

运行查询以获取过去 N 天的所有数据。
按天对这些记录进行分组
在过去 N 天的每一天将其合并到值为 { date => 0} 的哈希之上

【讨论】：

【解决方案2】：

好的，正如tpbowden 在他的回答中提到的那样，我设法将它减少到 1 个这样的查询：

date_format = "%d %b"
entries_last_week = time_entries.group_by { |t| t.created_at.to_date.strftime(date_format) }.map { |k, v| { k => v.map(&:duration_hours).sum } }
entries_hash = entries_last_week.reduce Hash.new, :merge
today = Date.today
days = (today - 6.days..today).map { |day| { day.strftime(date_format) => 0 } }
hash_of_days = days.reduce Hash.new, :merge

hash_of_days.merge(entries_hash)

【讨论】：