如何避免case case内的重复条件

Question

如何清理这一点？我想摆脱不必要的代码，像所有的重复a.question == f.question比较我每个case语句中使用：

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

记住红宝石让你在几个简单的传球做这些事情。有没有必要做这一切一气呵成：

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 

另一个速战速决是两个条款，有相同的代码相结合。

您可以通过在Answer模型上编写一个==方法来更轻松地进行比较。

Answer 2

您可以选择对的，只是a.question == f.question，然后将结果选择可以进一步选择使用。

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