SQLite、Ruby on Rails.如何使用 act_as_votable gem 获得最高投票对象? [英] SQLite, Ruby on rails. How to get the top voted objects with acts_as_votable gem?

问题描述

我正在尝试显示过去 7 天内最好的 3 个食谱.最好的方法取决于它使用acts_as_votable gem 获得的票数，但我还没有创建缓存.当前用户.rb，限制为过去 7 天

I"m trying to display the best 3 Recipes from the last 7 days. The best is dependant on the amount of votes it has using the acts_as_votable gem but i have not created a cache. current user.rb, limits to the last 7 days

def featuredfeed
    Recipe.where('created_at >= ?', Time.zone.now - 1.week)
  end

食谱.rb

class Recipe < ActiveRecord::Base

  belongs_to :user
  acts_as_votable

  ....

  default_scope -> { order(created_at: :desc) }

end

投票表已排序

id, votable_id, votable_type, voter_id, voter_type, vote_flag, vote_scope, vote_weight, created_at, updated_at

votable_id 是配方的 id，需要计算它被点赞的次数

The votable_id is the id of the recipe which needs to be counted for the number of times it has been upvoted

推荐答案

我想说，缓存投票是比下面一个更简洁的解决方案.这只是添加新迁移的问题.但取决于你..

I would like to say, that caching votes is much cleaner solution, than the following one. It is just a matter of adding a new migration. But up to you..

获取最近食谱的 id(这部分你已经完成了):

Get the ids of recent recipes (this part you already have done):

recipes_ids = Recipe.where('created_at >= ?', Time.zone.now - 1.week)

过滤此食谱的投票:

Vote
  .where(votable_id: recipes_ids)

按 votable_id 将它们分组:

  .group(:votable_id)

并找到那些拥有更多票的人，至少有 1 票:

And find those, which have more, at least 1 vote:

  .having('count(votable_id) > 1')

现在获取 3 个最受好评的食谱的 ID:

Now take the ids of 3 most upvoted recipes:

most_voted_recipes_ids = Vote
  .where(votable_id: recipes_ids)
  .group(:votable_id)
  .having('count(votable_id) > 1')
  .limit(3).votable_ids # or limit(3).pluck(:votable_id)

最后，找到那些最受欢迎的食谱:

And, lastly, find those most popular recipes:

most_popular_recipes = Recipe.where(id: most_voted_recipes_ids)

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