Ruby - 将滑轨的“业务逻辑”移动到服务模块中

Question

我想按照正确的OOP协议将一些滑轨业务逻辑转移到服务中。我遵循以下推荐的方法：Private module methods in Ruby

最终，我想将业务逻辑从模型和控制器转移到服务中，这些服务是简单的旧Ruby对象。因此，模型将只关注持久性，范围界定，验证。

这是制作服务模块的正确方法吗？

版本2：

module CategorizeJobs 

    def self.fetch(location, category_name) 
    c = Categorizer.new(location, category_name) 
    c.get_jobs 
    end 

    class Categorizer 

    def initialize(location, category_name) 
     @location = location 
     @category_name = category_name 
    end 

    def get_jobs 
     job_ids = get_correct_jobs 
     Category.includes(:jobs).where(jobs: { id: job_ids }) 
    end 

    private 

     def get_correct_jobs 
     jobs = filter_by_location 
     jobs = filter_by_category(jobs) 

     jobs.collect(&:id) 
     end 

     def filter_by_category(jobs) 
     return jobs unless @category_name.present? 

     category = Category.where(name: @category_name).first 
     if category 
      jobs = jobs.where(category: category) 
     end 

     jobs 
     end 

     def filter_by_location 
     if @location.present? 
      jobs = get_jobs_at_location 
     else 
      jobs = Job.open 
     end 
     end 

     def get_jobs_at_location(location) 
     Job.joins(:location).within(20, origin: @location).open 
     end 

    end 

end 

版本1：

module CategorizeJobs 

    def self.fetch(location, category_name) 
    c = Categorizer.new 
    c.perform(location, category_name) 
    end 


    class Categorizer 

    def perform(location, category_name) 
     job_ids = get_correct_jobs(location, category_name) 
     Category.includes(:jobs).where(jobs: { id: job_ids }) 
    end 

    private 

     def get_correct_jobs(location, category_name) 
     jobs = filter_by_location(location) 
     jobs = filter_by_category(jobs, category_name) 

     jobs.collect(&:id) 
     end 

     def filter_by_category(jobs, category_name) 
     return jobs unless category_name 

     category = Category.where(name: category_name).first 
     if category 
      jobs = jobs.where(category: category) 
     end 

     jobs 
     end 

     def filter_by_location(location) 
     if location 
      jobs = get_jobs_at_location(location) 
     else 
      jobs = Job.open 
     end 
     end 

     def get_jobs_at_location(location) 
     Job.joins(:location).within(20, origin: location).open 
     end 

    end 

end 

Answer 1

有一个宝石由Aldous的名称，可以帮助您实现服务对象的基础架构到您的Rails应用程序。看看它。

如果您需要关于如何使用它并构建您的应用程序的进一步帮助，您可以查看this和this获取更多帮助。