復雜的Rails / Postgres SQL優化

Question

餐廳has_many菜

Dish
has_many Photo

Photo
belongs_to Dish

Restaurant 1
  Dish 1
    Photo 1   May 9, 1:00 PM
  Dish 2
    Photo 2   May 9, 2:00 PM
  Dish 3
    Photo 3   May 9, 3:00 PM

Restaurant 2
  Dish 4
    Photo 4   May 9, 1:00 PM
  Dish 5
    Photo 5   May 9, 2:00 PM
  Dish 6
    Photo 6   May 9, 3:00 PM

我正在嘗試檢索最新的50張照片，每家餐廳最多只能有2張菜品照片。 鑒於以上數據，我將能夠檢索ID為2、3、5 2, 3, 5, and 6

至少可以說，我當前的實現很難看。

hash = {}
bucket = []
Photo.includes(:dish => [:restaurant]).order("created_at desc").each do |p|
  restaurant_id = p.dish.restaurant.id
  restaurant_count = hash[restaurant_id].present? ? hash[restaurant_id] : 0
  if restaurant_count < 2
    bucket << p
    hash[restaurant_id] = restaurant_count + 1
  end
  # if you've got 50 items short circuit.
end

我不禁感到有一個更有效的解決方案。 任何想法，將不勝感激 ：-）。

Answer 1

應該有一種對查詢進行“分組”的方法，但是至少以下內容要簡單一些：

def get_photo_bucket
  photo_bucket = restaurant_control = []
  Photos.includes(:dish => [:restaurant]).order("created_at desc").each do |photo|
    if photo_bucket.count < 50 && restaurant_control.count(photo.dish.restaurant.id) < 2
      photo_bucket << photo
      restaurant_control << photo.dish.restaurant.id
    end
  end
  photo_bucket
end