Question

我该如何清理一下？我想摆脱不必要的代码，就像我在每个a.question == f.question语句中使用的所有重复case比较一样：

def notifications_lookup(filters, answers)
    filters.flat_map do |f|
      answers.select do |a|
        case a.question.question_type
        when "image"
          a.question == f.question && a.answer_image.image_id == f.answer.to_i
        when "single"
          a.question == f.question && a.choice_answer.choice_id == f.answer.to_i
        when "list"
          a.question == f.question && a.choice_answer.choice_id == f.answer.to_i
        when "multiple"
          a.question == f.question && !(f.answer.split(",").map(&:to_i) & a.answer_multiple.choices.map(&:id)).empty?
        when "rating"
          results = verify_condition(f, a)
          a.question == f.question && results.any?
        else
          a.question == f.question
        end
      end
    end
  end

  def verify_condition(filter, a)
    a.answer_raitings.map do |r|
      r.raiting == filter.raiting && filter.answer.split(",").map(&:to_i).include?(r.response)
    end
  end

Answer 1

您可以只选择a.question == f.question，然后可以在进一步选择中使用该选择的结果。

def notifications_lookup(filters, answers)
    filters.flat_map do |f|
      selected_answers = answers.select {|a| a.question == f.question}
      selected_answers = selected_answers.select do |a|
        case a.question.question_type
        when "image"
          a.answer_image.image_id == f.answer.to_i
        when "single"
          a.choice_answer.choice_id == f.answer.to_i
        when "list"
          a.choice_answer.choice_id == f.answer.to_i
        when "multiple"
          !(f.answer.split(",").map(&:to_i) & a.answer_multiple.choices.map(&:id)).empty?
        when "rating"
          results = verify_condition(f, a)
          results.any?
        else
          true
        end
      end
    end
  end

Answer 2

记住Ruby允许你在几个简单的过程中做这些事情。没有必要一气呵成：

def notifications_lookup(filters, answers)
  filters.flat_map do |f|
    answers.select do |a|
      a.question == f.question
    end.select do |a|
      case a.question.question_type
      when "image"
        a.answer_image.image_id == f.answer.to_i
      when "single", "list"
        a.choice_answer.choice_id == f.answer.to_i
      when "multiple"
        !(f.answer.split(",").map(&:to_i) & a.answer_multiple.choices.map(&:id)).empty?
      when "rating"
        verify_condition(f, a).any?
      else
        true
      end
    end
  end
end

另一个快速解决方法是组合两个具有相同代码的子句。

您可以通过在答案模型上编写==方法，让您更轻松地进行比较。

如何避免case语句中的重复条件

2 个答案: