Has_many 嵌套形式不更新 db rails

Question

我正在建立一个博客类型的平台，用户可以在其中创建一个帐户，并且该用户有很多帖子，这些帖子应该显示在用户/显示页面上。我已经设置了帖子和用户表，并在事后添加了 belongs_to 和 has_many 关系。当我提交表单时，它会提醒“用户帐户已更新”，但帖子未显示在帖子表中。我想我一定错过了一些重要的东西。

Schema.rb

  create_table "posts", force: true do |t|
    t.text     "title"
    t.text     "body"
    t.string   "category1"
    t.string   "category2"
    t.datetime "created_at"
    t.datetime "updated_at"
    t.string   "image"
    t.integer  "user_id"
    t.integer  "category_id"
  end

  add_index "posts", ["category_id"], name: "index_posts_on_category_id", using: :btree
  add_index "posts", ["user_id"], name: "index_posts_on_user_id", using: :btree

  create_table "users", force: true do |t|
    t.string   "email"
    t.string   "password_digest"
    t.boolean  "admin"
    t.datetime "created_at"
    t.datetime "updated_at"
  end

型号

class User < ActiveRecord::Base
  has_many :posts
  accepts_nested_attributes_for :posts
end

class Post < ActiveRecord::Base
  belongs_to :user
  belongs_to :category
end

控制器

class UsersController < ApplicationController
  before_action :set_user, only: [:show, :edit, :update, :destroy]

  # GET /users
  # GET /users.json
  def index
    @users = User.all
  end

  def new
    @user = User.new
  end

  def create
    @user = User.new(user_params)

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

  # PATCH/PUT /users/1
  # PATCH/PUT /users/1.json
  def update
    respond_to do |format|
      if @user.update(user_params)
        format.html { redirect_to @user, notice: 'User was successfully updated.' }
        format.json { render :show, status: :ok, location: @user }
      else
        format.html { render :edit }
        format.json { render json: @user.errors, status: :unprocessable_entity }
      end
    end
  end

  # DELETE /users/1
  # DELETE /users/1.json
  def destroy
    @user.destroy
    respond_to do |format|
      format.html { redirect_to users_url, notice: 'User was successfully destroyed.' }
      format.json { head :no_content }
    end
  end

  private
    # Use callbacks to share common setup or constraints between actions.
    def set_user
      @user = User.find(params[:id])
    end

    # Never trust parameters from the scary internet, only allow the white list through.
    def user_params
      params.require(:user).permit(:email, :password, :password_confirmation)
    end
end

帖子/新的

<%= form_for current_user do |ff| %>
  <%= ff.fields_for :post do |f| %>
    <%= f.label :title %>
    <%= f.text_field :title, class: "title", placeholder: "Post Title..." %>

    <%= f.label :body %>
    <%= f.text_area :body, id: "edit" %>

    <%= f.submit %>
  <% end %>
<% end %>

Answer 1

Rails 4 在控制器中使用一种称为“强参数”的模式来确定可以批量分配对象的哪些属性。这是在用户控制器的底部：

# Never trust parameters from the scary internet, only allow the white list through.
def user_params
  params.require(:user).permit(:email, :password, :password_confirmation)
end

然后请注意，在create 操作中，这是在与@user = User.new(user_params) 在线调用的。由于您使用嵌套属性与用户一起创建帖子，因此您需要将帖子属性与允许的用户属性一起列入白名单，您可以通过传递哈希来完成，其中键是子对象的名称，值是需要为该对象列入白名单的属性数组。试试这个：

def user_params
  params.require(:user).permit(:email, :password, :password_confirmation, post: [:title, :body])
end

Answer 2

我通过去掉 fields_for 块并简化来解决这个问题：

# posts/new

<%= form_for @post do |f| %>
  <%= f.label :title %>
  <%= f.text_field :title, class: "title", placeholder: "Post Title..." %>

  <%= f.label :body %>
  <%= f.text_area :body, id: "edit" %>

  <%= f.submit %>
<% end %>

并将这一行添加到 posts 控制器的 create 方法中：

@post = Post.new post_params.merge(user_id: current_user.id)

current_user 是之前定义的辅助方法

Answer 3

您只允许存储用户参数，因为嵌套属性也允许发布参数。用以下代码替换您的 user_params 方法：

def user_params
  params.require(:user).permit(:email, :password, :password_confirmation, post_attributes: [:title, :body])
end