模型显示重构多个语言环境消息 rubocop

Question

当我写这个方法时，我知道它可以写得更好，但它是一个 MVP。现在，我正在尝试了解专家们认为应该采用的方法以及如何解决警察的问题。不断学习！

这里是有问题的控制器方法：

  def show
    if params[:id] && numeric?(params[:id])
      @job = Job.find_by(id: params[:id])
      if @job
        if @job.start_at
          if @job.end_at && @job.end_at >= Date.today
            @company = Company.find_by_id(@job.company_id)
            render 'jobs/show', layout: 'nested/job/show'
          else
            route_expired
          end
        else
          route_not_paid_for
        end
      else
        route_not_found_error
      end
    else
      route_not_found_error
    end
  end

然后这些调用一系列路由，根据他们获得的错误类型指导用户去哪里：

  private

  def route_not_found_error
    flash[:error] = t('flash_messages.jobs.id.not_found')
    redirect_to new_job_url
  end

  def route_expired
    flash[:error] = t('flash_messages.jobs.id.expired')
    redirect_to new_job_url
  end

  def route_not_paid_for
    flash[:error] = t('flash_messages.jobs.id.not_paid_for')
    redirect_to job_payment_url(@job)
  end

对于专家，您将如何处理这个问题？我知道这一定更干净。

Answer 1

首先，我会将一些逻辑移动到 Job 模型中

def started?
  self.start_at.present?
end

def finished?
  self.end_at.present?
end

def expired?
  finished? && self.end_at < Date.today
end

接下来，我认为不需要检查 :id 是否存在以及它是否为数字，您可以在控制器中使用 rescue_from 即可。综上所述，您可以按如下方式重构节目

class JobsController < ApplicationController
  rescue_from ActiveRecord::RecordNotFound, with: :route_not_found_error

  def show
    @job = Job.find params[:id]

    route_not_paid_for unless @job.started?
    route_expired if @job.expired?

    @company = Compandy.find_by_id(@job.company_id)
    render 'jobs/show', layout: 'nested/job/show'
  end

  private

  ...

end

模型显示重构多个语言环境消息 rubocop

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