Question

对Rails来说相当新，并且在这里尝试过很多东西而没有成功。

我的问题是当我使用这个嵌套表单发布到数据库时，我的一个表（apartment_images）发布了错误的外键（apartment_id）。

我有一个相当复杂的模型关系：我有一个构建 has_many_ through另一个表，它将（与其他）公寓表关联起来。有问题的 apartment_images 表属于公寓。

摘要版本如下：

建筑模型

has_many :building_relationships
has_many :apartments, :through => :building_relationships
accepts_nested_attributes_for :apartments, allow_destroy: true

公寓模型

belongs_to :building
has_many :apartment_images, -> { order(position: :asc) }, dependent: :destroy
has_many :building_relationships
has_many :buildings, :through => :building_relationships

ApartmentImage Model

belongs_to :apartment

buildings_controller（排除新方法）

def createNewBuilding
    @building = Building.new(building_params)
    @apartment = Apartment.where(building_id: @building.id) 
        #also tried this but results in no id being save:
         #@apartment = @building.apartments.build(apartment_params)

if @building.save
      redirect_to newBuilding_path, notice: "Successfully created building"
    else
      render 'newBuilding'#, notice: "ERROR"
    end


    if apartment_image_params
      apartment_image_params[:image].each do |value|
        @apartment.apartment_images.build({image: value}).save
            end
    end

  end

def apartment_image_params
      #also tried adding :apartment_id. didn't work.
      params.require(:apartment_image).permit(:id, image: []) if params[:apartment_image]
end

Answer 1

所以我终于开始工作了。不完全理解为什么这比我正在做的更有效，但是会对它进行阅读（但是，如果对你们这些人来说很明显，我也会全力以赴:)。

答案是从使用where / find / findby语句切换到使用Rails魔术：

def createNewBuilding
@building = Building.new(building_params)
@apartments = @building.apartments
@apartments.each do |apartments|

Rails嵌套表单创建错误的外键

1 个答案: