Question

我有四个模型：User，Product，Purchase和WatchedProduct。

我正在尝试获取符合以下条件之一的产品列表：

我创造了这个产品。
我买了这个产品。
该产品是免费的，我已经主演过＆＃34;或者＆＃34;观看＆＃34;它。

这是我到目前为止所做的：

class User < ActiveRecord::Base
  ...

  def special_products
    product_ids = []

    # products I created
    my_products = Product.where(:user_id => self.id).pluck(:id)
    # products I purchased
    my_purchases = Purchase.where(:buyer_id => self.id).pluck(:product_id)
    # free products I watched
    my_watched = WatchedProduct.where(:user_id =>self.id).joins(:product).where(products: { price: 0 }).pluck(:product_id)

    product_ids.append(my_products)
    product_ids.append(my_purchases)
    product_ids.append(my_watched)

    product_ids # yields something like this => [[1, 2], [], [2]]

    # i guess at this point i'd reduce the ids, then look them up again...

    product_ids.flatten!
    product_ids & product_ids
    products = []
    product_ids.each do |id|
      products.append(Product.find(id))
    end
    products

  end
end

我要做的是获取Product模型列表，而不是ID列表或ActiveRecord关系列表。我是新手，但有没有办法在一次连接中完成所有这些，而不是3次查询，减少和重新查找？

Answer 1

首先，我喜欢添加几个范围

class Product < ActiveRecord::Base
  scope :free, -> { where(price: 0) }
  scope :bought_or_created_by, lambda do |user_id|
    where('user_id = :id OR buyer_id = :id', id: user_id)
  end
end
class WatchedProduct < ActiveRecord::Base
  scope :by_user, ->(user_id) { where(user_id: user_id) }
end

然后是查询

special_products = (Product.bought_or_created_by(id) + Product.joins(:watched_products).free.merge(WatchedProduct.by_user(id)).uniq

这将使用2个查询返回一系列独特的产品。

Answer 2

虽然我不确定您的模型关联，但是您可以在单个查询中以某种方式执行所有这些操作：

Product.joins([{:user => :watched_products}, :buyer, :purchases]).where(["users.id = :current_buyer && buyers.id = :current_buyer && watched_products.user_id = :current_buyer && purchases.buyer_id = :current_buyer, products.price = 0", :current_buyer => self.id])

我假设

Product belongs_to user and buyer
Product has_many purchases
User has_many watched_products

Rails基于多个模型加入

2 个答案: