我有一个初始化方法，它做了一件愚蠢的事情。我需要将其优化为一个查询，但目前我的 SQL 技能让我失望了。我已经设想过使用 GROUP BY 和 UNION 以及各种各样的东西，但我只是让自己更加困惑。我将此遗赠给社区以提供一些见解:

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，他们将不胜感激! 谢谢。

最佳答案

我认为您无法在一个查询中干净利落地完成这项工作。 至少在不编写纯 sql 的情况下不会。

但是让我们尝试在 ActiveRecord 中找到一个好的解决方案

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')

可以重写

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

让我们将这个作用域保存为一个神奇的私有(private)方法 magic_scope

您当前的方法变为

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 方法可以通过对任一 load_stats_* 方法的结果求和得出。

如果 ActiveRecord 更聪明一点，我们就可以做到

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

这将在一次查询中为我们提供所需的所有数据。

但据我所知，目前这是不可能的。

但我希望这能让你更接近。

关于mysql - Rails 3 - 将多个计数转换为单个查询 - OrderedHash，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/6336188/