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我有 2 个模型

Class Ride
  has_many :trips

  #state (string: active or expired)
end

Class Trip
  #date (Date attribute)
  scope :active, -> (start_at = Date.today) { where("trips.date >= ?", [Date.today, start_at].max) }  
end

每天，我需要更新具有活动状态的游乐 设施的状态，所有行程的日期属性 < Date.today 如何在 1 个查询中执行此操作？我可以使用以下方法存档此类结果： 

Ride.with_active_state.select{|r| r.trips.active.size ==0}

但它使 huje 查询计算行程，eq：

    [1] pry(main)> Ride.with_active_state.select{|r| r.trips.active.size ==0}
   (7.3ms)  SELECT f_geometry_column,coord_dimension,srid,type FROM geometry_columns WHERE f_table_name='rides'
  Ride Load (1.6ms)  SELECT "rides".* FROM "rides" WHERE (rides.workflow_state = 'active')
   (2.9ms)  SELECT f_geometry_column,coord_dimension,srid,type FROM geometry_columns WHERE f_table_name='trips'
   (1.3ms)  SELECT COUNT(*) FROM "trips" WHERE "trips"."ride_id" = $1 AND (trips.date >= '2013-09-24')  [["ride_id", 9]]
   (0.7ms)  SELECT COUNT(*) FROM "trips" WHERE "trips"."ride_id" = $1 AND (trips.date >= '2013-09-24')  [["ride_id", 10]]
   (0.7ms)  SELECT COUNT(*) FROM "trips" WHERE "trips"."ride_id" = $1 AND (trips.date >= '2013-09-24')  [["ride_id", 11]]
   (0.7ms)  SELECT COUNT(*) FROM "trips" WHERE "trips"."ride_id" = $1 AND (trips.date >= '2013-09-24')  [["ride_id", 12]]
   (0.8ms)  SELECT COUNT(*) FROM "trips" WHERE "trips"."ride_id" = $1 AND (trips.date >= '2013-09-24')  [["ride_id", 13]]
   (0.8ms)  SELECT COUNT(*) FROM "trips" WHERE "trips"."ride_id" = $1 AND (trips.date >= '2013-09-24')  [["ride_id", 14]]
   (0.5ms)  SELECT COUNT(*) FROM "trips" WHERE "trips"."ride_id" = $1 AND (trips.date >= '2013-09-24')  [["ride_id", 15]]
   (0.5ms)  SELECT COUNT(*) FROM "trips" WHERE "trips"."ride_id" = $1 AND (trips.date >= '2013-09-24')  [["ride_id", 16]]
   (0.5ms)  SELECT COUNT(*) FROM "trips" WHERE "trips"."ride_id" = $1 AND (trips.date >= '2013-09-24')  [["ride_id", 17]]
   (0.5ms)  SELECT COUNT(*) FROM "trips" WHERE "trips"."ride_id" = $1 AND (trips.date >= '2013-09-24')  [["ride_id", 18]]
   (0.5ms)  SELECT COUNT(*) FROM "trips" WHERE "trips"."ride_id" = $1 AND (trips.date >= '2013-09-24')  [["ride_id", 19]]
   (0.5ms)  SELECT COUNT(*) FROM "trips" WHERE "trips"."ride_id" = $1 AND (trips.date >= '2013-09-24')  [["ride_id", 20]]

……

4

1 回答 1

2

Ride使用groupandhaving子句添加范围。它将检查骑行的所有未来行程的计数，并以 0 计数返回游乐设施。

Class Ride

  scope :active_state, where(state: "active")
  scope :with_nonactive_trips, -> (start_date = Date.today){ joins(:trips).
                                    group("rides.id").
                                    having( ["sum(trips.date > ?) = 0",start_date] ) }

end

Ride.active_state.with_nonactive_trips
# returns All the rides with state == active, alteast one trip and having no trips with date > Date.today

使用 alambda因为你在Trip. 我猜您需要使用与Date.today某些查询不同的日期。

于 2013-09-23T22:05:53.587 回答