Question

我正在寻找解决以下情况的最佳做法：

我有一个“添加剂”模型，它应该与其他一些模型相关联。

示例：

# Meal-Model
has_and_belongs_to_many :additives

# Offer-Model
has_and_belongs_to_many :additives

# Additive-Model
has_and_belongs_to_many :meals
has_and_belongs_to_many :meals

路线以下列方式嵌套：

resources :offers do
  resources :additives
end
resources :meals do
  resources :additives
end

所以我得到这样的网址：

/offers/123/additives
/meals/567/additives

两条路线都会导致相同的控制器操作，即additives#index。在AdditivesController中，我检查params是否可用于选择要获取的数据：

class AdditivesController < ApplicationController

before_filter :offermealswitch

# GET /restaurants/1/meals/123/additives
# GET /restaurants/1/offers/123/additives
def index   
  @additives = @additivemeal.additives    
end

def offermealswitch
  if params.has_key?(:meal_id)
    @additivemeal = Meal.find(params[:meal_id])
    @type = "Meal"
  elsif params.has_key?(:offer_id)
    @additivemeal = Offer.find(params[:offer_id])
    @type = "Offer"
  end
end

end

这是处理这个问题的正确方法吗？它工作得很好，但我不是shure这是轨道方式...... 谢谢你的回答！

Answer 1

关于@Taryn East的

编辑

resources :offers do
  resources :additives, :type => "Offer"
end
resources :meals do
  resources :additives, :type => "Meal"
end

class AdditivesController < ApplicationController
  before_filter :find_additive

  def index   
    @additives = @additive.additives    
  end

  private
  def find_additive
    @type = params[:type]
    @additive = @type.constantize.find([@type, "id"].join("_")) # or "#{@type}_id", as you wish
  end
end

Answer 2

叹息切换到应答空间，这样我至少可以添加回车并使代码不会变得愚蠢。

我同意fl00r的回答，但会补充说你需要实例化对象：

@type = params[:type] 
@obj = @type.constantize.find(params["#{type}_id"])
@additives = @obj.additives

Rails：根据路由不同的控制器行为

2 个答案: