只渲染每个酒店价格最低的房间

Question

我正在从我的数据库中检索一组特定的房间。一个房间belongs_to一个酒店。我只检索属于靠近请求区域（20 公里）的酒店的房间，我还在检查房间是否有合适的容量。这为我提供了一组特定的房间（有时每家酒店都有很多房间）。我只想渲染一个符合每家酒店标准的房间：最便宜的房间 room_price。我该怎么做？

到目前为止，我的方法是这样的

  def find_hotels
    # hotels near query
    @hotels = Hotel.near(params[:place], 20)
    number_of_people = params[:adults_number].to_i + params[:children_number].to_i
    # rooms with the right capaciy
    # rooms with the best price
    @rooms = Room.where(hotel_id: @hotels.map(&:id)).where("capacity >= :number_of_people", {number_of_people: number_of_people})
  end

Answer 1

这个呢？

def find_hotels
    # hotels near query
    @hotels = Hotel.near(params[:place], 20)
    number_of_people = params[:adults_number].to_i + params[:children_number].to_i
    # rooms with the right capaciy
    # rooms with the best price
    list_of_rooms = @hotels.inject({}){|result, hotel| result[hotel.id] = cheapest_room_id(hotel.id, number_of_people); result}
  end

def cheapest_room_id(hotel_id, number_of_people)
  return Room.where(hotel_id: hotel_id).where("capacity > ?", number_of_people).order("room_price ASC").first.id
end

变量 list_of_rooms 将包含以下形式的散列：

{ hotel_1 => room_123, hotel_2 => room_44, hotel_3 => room_666 }

这里的一切都是一个ID。

PS：应该可以。

Answer 2

我猜你可能会考虑酒店可能会有不止一间最低价相同的房间。

无论如何，如果您要考虑 100 家酒店，那么您可能会发现运行对每个酒店进行一次查询以查找最便宜的房间是不可接受的。

如果是这样，您可能需要深入研究 SQL 来优化搜索（顺便说一句，您也可以通过组合查找酒店的查询和查找房间的查询来进行优化）。

像这样的东西应该是高性能的。

def find_hotels
  # hotels near query
  number_of_people = params[:adults_number].to_i + params[:children_number].to_i
  # rooms with the right capaciy
  # rooms with the best price
  @rooms = Room.where(hotel: Hotel.near(params[:place], 20)).
                where("capacity >= :number_of_people", {number_of_people: number_of_people}).
                where("not exists (select null
                                     from rooms r2
                                    where r2.hotel_id   =  rooms.hotel_id    and
                                          r2.capacity   >= :number_of_people and
                                          r2.room_price <= rooms.room_price  and
                                          r2.id         <= rooms.id)", , {number_of_people: number_of_people})
end

它找到了在同一家酒店中没有其他房间的房间，并且具有所需的容量和更便宜的价格。事实上，假设您只希望每个酒店返回一个房间，它会更进一步。

如果您想要以最便宜的价格退回所有房间，请使用：

def find_hotels
  # hotels near query
  number_of_people = params[:adults_number].to_i + params[:children_number].to_i
  # rooms with the right capaciy
  # rooms with the best price
  @rooms = Room.where(hotel: Hotel.near(params[:place], 20)).
                where("capacity >= :number_of_people", {number_of_people: number_of_people}).
                where("not exists (select null
                                     from rooms r2
                                    where r2.hotel_id   =  rooms.hotel_id    and
                                          r2.capacity   >= :number_of_people and
                                          r2.room_price < rooms.room_price)", , {number_of_people: number_of_people})
end

只渲染每个酒店价格最低的房间

Only render rooms with the cheapest price for each hotel

activerecord

ruby-on-rails

querying