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mysql-Rails 3-将多个计数转换为单个查询-OrderedHash

2019-12-08 09:17:26 阅读：231 来源： 互联网

标签：activerecord optimization count ruby-on-rails mysql

我有一个做傻事的初始化方法.我需要将其优化为一个查询,但是我的SQL技能目前使我感到失望.我已经考虑过使用GROUP BY和UNION以及各种各样的东西,但是我只是让自己更加困惑.我将其遗赠给社区,以得出一些见解：

Class Stats
  # Turn these three queries into one query that we can then
  # load into the three different instance variables
  def initialize(question)
    # Integer = total number of answers for this question
    @total = total_answers(question.id)

    # Hash keyed by 0 (incorrect answers) and 1 (correct answers)
    @stats_total = load_stats_total(question.id) if @total > 0

    # Hash keyed by answer_id with values = total number of answers
    @stats_answers = load_stats_answers(question.id) if @total > 0
  end

  # Returns an int = the total number of answer attempts for
  # this question (right + wrong user_answers)
  # Excludes anonymous users
  def total_answers(question_id)
    UserAnswer.count(:conditions => ['u.anonymous = 0 AND q.id = ?', question_id], :joins => 'JOIN answers a ON user_answers.answer_id = a.id JOIN questions q ON q.id = a.question_id JOIN users u ON u.id = user_answers.user_id')
  end

  # Returns an OrderedHash =
  # {"0" => number of wrong user_answers for this question,
  #  "1" => number of correct user_answers for this question}
  # Excludes anonymous users
  def load_stats_total(question_id)
    UserAnswer.count(:conditions => ['u.anonymous = 0 AND q.id = ?', question_id], :joins => 'JOIN answers a ON user_answers.answer_id = a.id JOIN questions q ON q.id = a.question_id JOIN users u ON u.id = user_answers.user_id', :group => 'a.correct')
  end

  # Returns an OrderedHash =
  # {
  #  some_answer_id => total number of user_answers for this answer,
  #  some_other_answer_id => total number of user_answers for this answer
  #  ...
  # }
  # Excludes anonymous users
  def load_stats_answers(question_id)
    UserAnswer.count(:conditions => ['u.anonymous = 0 AND q.id = ?', question_id], :joins => 'JOIN answers a ON user_answers.answer_id = a.id JOIN questions q ON q.id = a.question_id JOIN users u ON u.id = user_answers.user_id', :group => 'a.id')
  end
end

如果有人有任何想法,将不胜感激！
谢谢.

解决方法:

我认为您无法在一个查询中做到这一点.
至少并非没有编写纯sql.

但是,让我们尝试在ActiveRecord中找到一个不错的解决方案

首先,让我们尝试删除一些sql

UserAnswer.count(:conditions => ['u.anonymous = 0 AND q.id = ?', question_id], :joins => 'JOIN answers a ON user_answers.answer_id = a.id JOIN questions q ON q.id = a.question_id JOIN users u ON u.id = user_answers.user_id')

可以改写

UserAnswer.joins(:user).where(:users => {:anonymous => false})\
  .joins(:answer => :question).where(:questions => {:id => question_id})\
  .count

让我们将范围保存为魔术私有方法magic_scope

您当前的方法成为

def total_answers(question_id)
  magic_scope(question_id).count
end

def load_stats_total(question_id)
  magic_scope(question_id).count(:group => "answers.correct")
end

def load_stats_answers(question_id)
  magic_scope(question_id).count(:group => "answers.id")
end

当然,值得注意的是,total_answers方法可以通过汇总两个load_stats_ *方法的结果得出.

如果ActiveRecord更加聪明,我们可以做

def all_the_magic(question_id)
  magic_scope(question_id).count(:group => ["answers.correct", "answers.id"])
end

这样一来,我们就能获得所需的所有数据.

但据我所知,目前尚不可能.

但我希望这能使您更接近.

标签：activerecord,optimization,count,ruby-on-rails,mysql
来源： https://codeday.me/bug/20191208/2090355.html

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mysql-Rails 3-将多个计数转换为单个查询-OrderedHash