通过关系创建 has_one

Question

在我的预订表格中，我想创建一个用户可以预订一个房间的情况。由于我需要为预订的房间提供唯一标识符，因此我使用了通过reservation_rooms 链接房间和预订的设置。

• reservation_room 属于reservation 和room
• 一个reservation has_one room

问题

我试图让用户选择一个房间，然后用这个reservation 和room picked 构建reservation_room 模型

我目前的设置不允许我首先获得房间（在创建操作中）：

Processing by ReservationsController#create as HTML
  Parameters: {"utf8"=>"✓", "authenticity_token"=>"YfVsfb4po0Dl/kycf9Mm6LMYCXi7+GLDUpe8gANaGRMNzPiCoq6HC9CWarSyJjLmYG1/hE6D7w==", "reservation"=>{"arrival"=>"2019-12-05", "departure"=>"2019-12-06", "room_id"=>"7"}, "commit"=>"Search", "hotel_id"=>"22"}

Room.find(params[:room_id])
ActiveRecord::RecordNotFound: Couldn't find Room without an ID

代码

型号

models
class Reservation < ApplicationRecord
  belongs_to :hotel
  has_one :reservation_room, dependent: :destroy
  has_one :room, through: :reservation_rooms
end

class ReservationRoom < ApplicationRecord
  belongs_to :room
  belongs_to :reservation
end

class Room < ApplicationRecord
  belongs_to :room_category
  has_many :reservation_rooms
  has_many :reservations, through: :reservation_rooms, dependent: :destroy
  accepts_nested_attributes_for :room_category
end

reservations_controller

def new
    @hotel = Hotel.find(params[:hotel_id])
    @reservation = @hotel.reservations.new
    @room_categories = @hotel.room_categories
    @rooms = Room.where(room_category: @room_categories)
    authorize @reservation
  end

  def create
    @hotel = Hotel.find(params[:hotel_id])
    @reservation = @hotel.reservations.new(reservation_params)
    binding.pry
    @room = Room.find(params[:room_id])
    @reservation.build_reservation_room(room_id: @room.id)
    authorize @reservation
    if @reservation.save
      authorize @reservation
      redirect_to new_second_part_hotel_reservation_path(@hotel, @reservation)
    end
  end

  private

  def reservation_params
      params.require(:reservation).permit(:room_id, :arrival, :departure,
      reservation_room_attributes: [:room_id, :reservation_id],
        rooms_attributes: [:id,:name, :room_category_id,
          room_categories_attributes: [:id, :name]])
  end
end

reservations/new.html.erb

  <%= simple_form_for [@hotel, @reservation] do |f|%>
        <%= f.input :arrival%>
        <%= f.input :departure %>
        <%= f.input :room_id, collection: @room_categories.order(:name), as: :grouped_select, group_method: :rooms,  label:false %>
        <%= f.button :submit %>
    <% end %>

Answer 1

如果您有一个多对多关联（预订可以预订多个房间），那么使用连接表是有意义的，但如果您只想要一对多关联，您只想在预订上放置一个外键并使用 @987654321 @

rails g migration add_room_to_reservations room:belongs_to

class Reservation
  belongs_to :room
end

class Room
  has_many :reservations
end

恕我直言，每个房间预订一次可能更明智。