Rails 选择 GROUP 中 COUNT 最高的对象

目标是选择Store其中一个Coupon最常用于.

目前，我有这个，并且它有效（分解以供解释）：

# coupon.rb
  has_many :redemptions
  has_and_belongs_to_many :stores

  def most_popular_store
    stores.find(        # Return a store object
      redemptions       # Start with all of the coupon's redemptions
      .group(:store_id) # Group them by the store_id
      .count            # Get a hash of { 'store_id' => 'count' } values
      .keys             # Create an array of keys
      .sort             # Sort the keys so highest is first
      .first            # Take the ID of the first key
    )
  end
###

它的用法如下：

describe 'most_popular_store' do
    it 'returns the most popular store' do
        # Create coupon
        coupon = FactoryGirl.create(:coupon)

        # Create two different stores
        most_popular_store = FactoryGirl.create(:store, coupons: [coupon])
        other_store        = FactoryGirl.create(:store, coupons: [coupon])

        # Add redemptions between those stores
        FactoryGirl.create_list(:redemption, 2, coupon: coupon, store: other_store)
        FactoryGirl.create_list(:redemption, 5, coupon: coupon, store: most_popular_store)

        # Verify
        expect(coupon.most_popular_store.title).to eq most_popular_store.title
    end
end

就像我说的，这个方法是有效的，但它看起来像是经过猴子补丁的。我怎样才能重构我的most_popular_store method?

我觉得你的方法其实行不通。count为您提供一个哈希值，其中键为 store_ids，值作为计数，然后运行keys在哈希上，它为您提供了一个 store_ids 数组。从那时起，您就失去了计数，您将按 store_ids 排序并获取第一个。您的测试通过的唯一原因是您在另一个商店之前创建了受欢迎的商店，因此它的 ID 较低（sort默认情况下按升序排序）。要获得正确的结果，请进行以下更改：

redemptions       # Start with all of the coupon's redemptions
  .group(:store_id) # Group them by the store_id
  .count            # Get a hash of { 'store_id' => 'count' } values
  .max_by{|k,v| v}  # Get key, val pair with the highest value
                    # output => [key, value]
  .first            # Get the first item in array (the key)