我应该重构这些嵌套的每个块吗？

Question

我想知道是否有一种方法来重构此方法以最大限度地缩减缩进级别。读完干净的代码后，鲍勃叔叔说，一旦方法超过两个级别，就很难理解正在发生的事情。我不认为这种方法过于复杂，但我认为它可以被清理，看起来更像一个故事。

这里是我的方法

def sync!(xml_document, language_id, status_id, file) 
    survey = sync_survey.(xml_document, status_id) 
    sync_survey_label.(xml_document, survey.id, language_id, file) 

    xml_document.xpath('//page').each do |xml_section| 
    section = sync_section.(xml_section, survey.id) 
    sync_section_label.(xml_section, section.id, language_id) 

    xml_section.xpath('.//question_group').each do |xml_question_group| 
     question_group = sync_question_group.(xml_question_group, section.id) 
     sync_question_group_label.(xml_question_group, question_group.id, language_id) 

     xml_question_group.xpath('.//question').each do |xml_question| 
     question = sync_question.(xml_question, question_group.id) 
     sync_question_label.(xml_question, question.id, language_id) 

     xml_question.xpath('.//answerchoice').each do |xml_choice| 
      choice = sync_choice.(xml_choice, question.id) 
      sync_choice_label.(xml_choice, choice.id, language_id) 
     end 
     end 
    end 
    end 

    survey 
end 

所以，林遍历XML并保存到相应的方法。每个调查都有很多部分，其中有很多问题组有很多选择的问题。

我知道我可以提取每个循环到一个方法，但我仍然会有4个级别的缩进。

我想要做这样的事情......

def sync!(xml_document, language_id, status_id, file) 
    survey = sync_survey.(xml_document, status_id) 
    sync_survey_label.(xml_document, survey.id, language_id, file) 
    sync_sections.(xml_document) 
    sync_section_labels.(xml_document, language_id) 
    sync_question_groups.(xml_document) 
    sync_question_group_labels.(xml_document, language_id) 
    sync_questions.(xml_document) 
    sync_question_labels.(xml_document, language_id) 
    sync_choices.(xml_document) 
    sync_choice_labels.(xml_document, language_id) 

    survey 
end 

是否有任何想法，做这样的事情还是让这段代码更易读？是否有必要重构它？任何想法都欢迎。

这里是满级

class SurveyBuilder 
    attr_reader :xml_parser, :sync_survey, :sync_survey_label, :sync_section, :sync_section_label, :sync_question, :sync_question_label, :sync_question_group, :sync_question_group_label, :sync_choice, :sync_choice_label 

    def initialize 
    @sync_survey = SurveyBuilder::Sync::Survey.new 
    @sync_survey_label = SurveyBuilder::Sync::SurveyLabel.new 
    @sync_section = SurveyBuilder::Sync::Section.new 
    @sync_section_label = SurveyBuilder::Sync::SectionLabel.new 
    @sync_question = SurveyBuilder::Sync::Question.new 
    @sync_question_label = SurveyBuilder::Sync::QuestionLabel.new 
    @sync_question_group = SurveyBuilder::Sync::QuestionGroup.new 
    @sync_question_group_label = SurveyBuilder::Sync::QuestionGroupLabel.new 
    @sync_choice = SurveyBuilder::Sync::Choice.new 
    @sync_choice_label = SurveyBuilder::Sync::ChoiceLabel.new 
    @xml_parser = SurveyBuilder::XMLParser.new 
    end 

    def call(file, status_id) 
    xml_document = xml_parser.(file) 
    language_id = Language.find_by(code: xml_document.xpath('//lang_code').children.first.to_s).id 

    survey = sync!(xml_document, language_id, status_id, file) 

    survey 
    end 

private 
    def sync!(xml_document, language_id, status_id, file) 
    survey = sync_survey.(xml_document, status_id) 
    sync_survey_label.(xml_document, survey.id, language_id, file) 

    xml_document.xpath('//page').each do |xml_section| 
     section = sync_section.(xml_section, survey.id) 
     sync_section_label.(xml_section, section.id, language_id) 

     xml_section.xpath('.//question_group').each do |xml_question_group| 
     question_group = sync_question_group.(xml_question_group, section.id) 
     sync_question_group_label.(xml_question_group, question_group.id, language_id) 

     xml_question_group.xpath('.//question').each do |xml_question| 
      question = sync_question.(xml_question, question_group.id) 
      sync_question_label.(xml_question, question.id, language_id) 

      xml_question.xpath('.//answerchoice').each do |xml_choice| 
      choice = sync_choice.(xml_choice, question.id) 
      sync_choice_label.(xml_choice, choice.id, language_id) 
      end 
     end 
     end 
    end 
    survey 
    end 

end 

Answer 1

我的第一个尝试会是这样的：

def sync!(xml_document, language_id, status_id, file) 
    survey = sync_survey.(xml_document, status_id) 
    sync_survey_label.(xml_document, survey.id, language_id, file) 

    xml_document.xpath('//page').each do |xml_section| 
    sync_section(xml_section, ...) 
    end 

    survey 
end 

def sync_section(xml_section, ...) 
    ... 
    xml_section.xpath('.//question_group').each do |xml_question_group| 
    sync_question_group(xml_question_group, ...) 
    end 
end 

def sync_question_group(xml_question_group, ...) 
    ... 
    xml_question_group.xpath('.//question').each do |xml_question| 
    sync_question(xml_question, ...) 
    end 
end 

def sync_question(xml_question, ...) 
    ... 
    xml_question.xpath('.//answerchoice').each do |xml_choice| 
    sync_choice(xml_choice, ...) 
    end 
end 

def sync_choice(xml_choice, ...) 
    ... 
end 

我省略的公平一点，因为我不知道你是如何定义的现有sync_question_group所使用以一种令人困惑的方式使用点语法。