Rails – 保存现有记录 ID 而不是复制记录

Question

我现在正在学习 Rails，我是个新手。我希望能够在这里得到 Rails 专家的帮助。我尝试通过 StackOverflow 进行搜索，但未能找到解决方案。

所以有一个 Pet 模型，它的引用有 Breed 模型，它引用了 Category 模型，所以 3 个模型：

class Pet < ApplicationRecord
  belongs_to :breed
  accepts_nested_attributes_for :breed
  has_one :category, through: :breed

class Breed < ApplicationRecord
  belongs_to :category
  has_many :pets, dependent: :destroy
  accepts_nested_attributes_for :category

class Category < ApplicationRecord
  has_many :breeds, dependent: :destroy
  has_many :pets, through: :breed

我已经使用simple_form 设置了表单。创建新宠物时，用户可以select 与Breed 关联的现有Category 选项。然后在表单中，将breed.name 属性设置为text_field，用户可以输入自己的。请在嵌套属性部分下方：

...
<!-- Nested Breed & Category Attributes -->
<%= f.simple_fields_for :breed, Breed.new do |breed| %>
  <%= breed.error_notification %>
  <%= breed.error_notification message: breed.object.errors[:base].to_sentence if breed.object.errors[:base].present? %>
  <div class="form-row">

    <div class="form-group col-md-4">
      <%= breed.association :category, collection: Category.order(:id),
                                       label: "Animal Category", required: true %>
    </div>

    <div class="form-group col-md-8">
      <%= breed.input :name, required: true,
                             label: "Breed", placeholder: "Domestic short hair",
                             hint: "Type 'unknown', if unsure", class: "form-control" %>
    </div>
  </div>
<% end %>

所以我的问题是，我怎样才能用现有的breed_id 记录保存新宠物而不用不同的breed_id 复制它？

所以目前，我已经有一个现有的记录 breed_id : 1，它由 category_id: 1 (for Cat), and name: 'domestic short hair' 组成——如果我在表单中创建一个具有这些相同值的新宠物，它将复制记录，它应该保存breed_id: 1，但是新宠物却保存在breed_id: 15下...

当我查看console:

irb(main):001:0> Breed.find_by_name("domestic short hair")
  Breed Load (6.8ms)  SELECT "breeds".* FROM "breeds" WHERE "breeds"."name" = $1 LIMIT $2  [["name", "domestic short hair"], ["LIMIT", 1]]
=> #<Breed id: 1, name: "domestic short hair", category_id: 1, created_at: "2021-08-14 13:12:41.266154000 +0000", updated_at: "2021-08-14 13:12:41.266154000 +0000">
irb(main):002:0> Breed.last
  Breed Load (8.2ms)  SELECT "breeds".* FROM "breeds" ORDER BY "breeds"."id" DESC LIMIT $1  [["LIMIT", 1]]
=> #<Breed id: 15, name: "domestic short hair", category_id: 1, created_at: "2021-08-16 04:08:31.602249000 +0000", updated_at: "2021-08-16 04:08:31.602249000 +0000">

更新我的controller / model / form helper（我不确定是哪一个）的最佳方法是什么，这样可以防止重复？我目前没有breeds_controller，只使用pets_controller，这是Scaffolding 的一个基本的：

pets_controller.rb

  def show
  end

  # GET /pets/new
  def new
    @pet = Pet.new
    @pet.build_breed
  end
  
  # GET /pets/1/edit
  def edit
  end
  
  # POST /pets or /pets.json
  def create
    @pet = current_user.pets.new(pet_params)

    respond_to do |format|
      if @pet.save
        format.html { redirect_to @pet, notice: "Pet was successfully added." }
        format.json { render :show, status: :created, location: @pet }
      else
        format.html { render :new, status: :unprocessable_entity }
        format.json { render json: @pet.errors, status: :unprocessable_entity }
      end
    end
  end

...

  private

  def pet_params
   params.require(:pet).permit(:owner_id, :name, :dob, :gender, :bio, :instagram,
                                breed_attributes: [:name, :category_id])
  end

我尝试在模型breed.db中写一个before_save :existing_breed_record，我认为这是错误的......

class Breed < ApplicationRecord
  belongs_to :category
  has_many :pets, dependent: :destroy 
  accepts_nested_attributes_for :category
  before_save :existing_breed_record

  private

    def existing_breed_record
     if self.find_by(category_id, name).exists?
      return self.id
    end
end

这是我的schema.db

  create_table "breeds", force: :cascade do |t|
    t.string "name"
    t.bigint "category_id", null: false
    t.datetime "created_at", precision: 6, null: false
    t.datetime "updated_at", precision: 6, null: false
    t.index ["category_id"], name: "index_breeds_on_category_id"
  end

  create_table "categories", force: :cascade do |t|
    t.string "name"
    t.datetime "created_at", precision: 6, null: false
    t.datetime "updated_at", precision: 6, null: false
  end

  create_table "pets", force: :cascade do |t|
    t.bigint "owner_id", null: false
    t.bigint "breed_id", null: false
    t.string "name"
    t.date "dob"
    t.integer "gender"
    t.string "bio"
    t.string "instagram"
    t.datetime "created_at", precision: 6, null: false
    t.datetime "updated_at", precision: 6, null: false
    t.index ["breed_id"], name: "index_pets_on_breed_id"
    t.index ["owner_id"], name: "index_pets_on_owner_id"
  end

我从指南中读到有 find_or_create_by 方法，我认为这是我需要的，但我不确定在哪里以及如何实现它......

对于数据库和 Rails 专家，为了学习，您建议我下次如何做得更好？我是否应该将 breed 作为字符串属性/列添加到 Pet 模型，并直接将 Category 模型与 Pet 关联？

Answer 1

如果你想使用嵌套形式，你可以将before_save方法添加到Pet

\\pet.rb

before_save :check_existing_breed


def check_existing_breed
  if self.breed&.name and breed = Breed.find_by(name: self.breed&.name)
    self.breed = breed
  end
end

Answer 2

在老师的帮助下，我们终于找到了解决办法。有点乱，但它不再重复任何现有记录！

首先，感谢@Yunwei.W，按照建议，我在嵌套形式中添加了@pet.breed（之前是:breed, Pet.new do ...：

...
<!-- Nested Breed & Category Attributes -->
<%= f.simple_fields_for :breed, @pet.breed do |breed| %>
...

然后在pets_controller.rb：

def create
 # Preventing duplicates when breed name mathes exsiting record
 breed = Breed.find_by_name(params[:pet][:breed_attributes][:name]) || Breed.create(name: params[:pet][:breed_attributes][:name], category_id: params[:pet][:breed_attributes][:category_id])

 @pet = current_user.pets.new(pet_params.except(params[:pet][:breed_attributes][:name]))

 @pet.breed = breed

...

end

肯定有比这更好的解决方案，但我很高兴在使用此阻止程序 3 天后它终于起作用了。感谢所有为帮助做出贡献的人！真的很感激！