如何正确创建和更新具有活动记录关联的帖子？

Question

所以我遇到了这个挑战，我无法弄清楚如何声明控制器以使其正常工作，我遇到了各种各样的错误，但我的主要问题是，现在我可以'不保存新配方或更新与下拉菜单中使用的 recipe_type 模型关联的 'Tipo de Receita'，请记住，recipe_type.name 将预先填充

这是进行编辑的edit.hmtl.erb

<%= form_for @recipe do |f| %>
  <%= f.label :title, 'Título' %>
  <%= f.text_field :title %>
  <%= f.label :recipe_type, 'Tipo da Receita' %>
  <%= collection_select(:recipe, :recipe_type_id, RecipeType.all, :id, :name) %>
  <%= f.label :cuisine, 'Cozinha' %>
  <%= f.text_field :cuisine %>
  <%= f.label :difficulty, 'Dificuldade' %>
  <%= f.text_field :difficulty %>
  <%= f.label :cook_time, 'Tempo de Preparo' %>
  <%= f.number_field :cook_time %>
  <%= f.label :ingredients, 'Ingredientes' %>
  <%= f.text_area :ingredients %>
  <%= f.label :cook_method, 'Como Preparar' %>
  <%= f.text_area :cook_method %>
  <%= f.submit 'Enviar' %>
<% end %>

同样是注册新配方的 new.hmtl.erb

<%= form_for @recipe do |f| %>
  <%= f.label :title, 'Título' %>
  <%= f.text_field :title %>
  <%= f.label :recipe_type, 'Tipo da Receita' %>
  <%= collection_select(:recipe, :recipe_type_id, RecipeType.all, :id, :name) %>
  <%= f.label :cuisine, 'Cozinha' %>
  <%= f.text_field :cuisine %>
  <%= f.label :difficulty, 'Dificuldade' %>
  <%= f.text_field :difficulty %>
  <%= f.label :cook_time, 'Tempo de Preparo' %>
  <%= f.number_field :cook_time %>
  <%= f.label :ingredients, 'Ingredientes' %>
  <%= f.text_area :ingredients %>
  <%= f.label :cook_method, 'Como Preparar' %>
  <%= f.text_area :cook_method %>
  <%= f.submit 'Enviar' %>
<% end %>

models/recipe_type 这是下拉菜单中使用的一种

class RecipeType < ApplicationRecord
    has_many :recipes
    validates :name, presence: true
end

models/recipe 接收 recipe_type 模型

class Recipe < ApplicationRecord
  belongs_to :recipe_type
  validates :title, :cuisine, :difficulty, :cook_time,
            :ingredients, :cook_method, presence: true

  def cook_time_min
    "#{cook_time} minutos"
  end
end

recipes.controller.rb

class RecipesController < ApplicationController
  def index
    @recipes = Recipe.all
  end

  def show
    @recipe = Recipe.find(params[:id])
  end

  def new
    @recipe = Recipe.new
    @recipe_type = RecipeType.all
  end

  def create
    @recipe_type = RecipeType.all
    @recipe = Recipe.new(recipe_params)
    if @recipe.save
      redirect_to @recipe
    else
      flash[:alert] = 'Você deve informar todos os dados da receita'
      render :new
    end
  end

  def edit
    @recipe_type = RecipeType.all
    @recipe = Recipe.find(params[:id])
    @recipe_type = RecipeType.all
  end

  def update
    @recipe = Recipe.find(params[:id])
    if @recipe.update(recipe_params)
      redirect_to @recipe
    else
      flash[:alert] = 'Você deve informar todos os dados da receita'
      render :edit
    end
  end

  private

  def recipe_params
    params.require(:recipe).permit(:title, :cuisine, :difficulty,
                                   :cook_time, :ingredients, :cook_method, :name)
  end

end

这就是架构

ActiveRecord::Schema.define(version: 2020_03_26_013134) do

  create_table "recipe_types", force: :cascade do |t|
    t.string "name"
    t.integer "recipe_id"
    t.datetime "created_at", null: false
    t.datetime "updated_at", null: false
    t.index ["recipe_id"], name: "index_recipe_types_on_recipe_id"
  end

  create_table "recipes", force: :cascade do |t|
    t.string "title"
    t.string "cuisine"
    t.string "difficulty"
    t.integer "cook_time"
    t.datetime "created_at", null: false
    t.datetime "updated_at", null: false
    t.text "ingredients"
    t.text "cook_method"
    t.integer "recipe_type_id"
    t.index ["recipe_type_id"], name: "index_recipes_on_recipe_type_id"
  end
end

Answer 1

一个问题是您没有从表单构建器构建输入。你想使用：

<%= f.label :recipe_type_id, 'Tipo da Receita' # key has to match input for accessibility %>
<%= f.collection_select(:recipe_type_id, RecipeType.all, :id, :name) %>

collection_select 是一个简单的输入助手，它只会创建一个选择标签，该标签未绑定到表单构建器，因此未绑定到您的模型实例。因此，如果记录无效或您正在更新现有记录，则不会选择任何内容。

f.collection_select 是表单构建器上的一个方法，将正确设置表单包装的模型实例中的recipe_type_id 值。

您还需要允许白名单中的参数：

def recipe_params
  params.require(:recipe)
        .permit(
          :title, :cuisine, :difficulty,
          :cook_time, :ingredients, :cook_method, 
          :name, :recipe_type_id
        )
end

Answer 2

您需要在recipe_params中添加recipe_type_id：

def recipe_params
    params.require(:recipe).permit(:recipe_type_id ...)
end