Question

嗨，我有一个应用程序，它使用devise进行身份验证并设计为invitable。

注册后，用户将创建一个帐户。

class Account < ApplicationRecord
  belongs_to :user, class_name: "owner", foreign_key: "owner_id"
  has_many :users, dependent: :destroy
  has_many :clients, dependent: :destroy
end

用户注册并在创建时默认获得admin角色！

    class User < ApplicationRecord
  has_merit
  enum role: [:user, :tech, :admin, :manager]
  has_one :account, foreign_key: 'owner_id'
  accepts_nested_attributes_for :account
  after_initialize :set_default_role, :if => :new_record?

  def set_default_role
    self.role ||= :admin
  end

  # Include default devise modules. Others available are:
  # :confirmable, :lockable, :timeoutable, :trackable and :omniauthable
  devise :database_authenticatable, :invitable, :registerable,
         :recoverable, :rememberable, :validatable
end

我对如何管理用户has_one：作为所有者的帐户（用户注册）和belongs_to：作为雇员的帐户（邀请用户）感到困惑

模式

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

create_table "users", force: :cascade do |t|
    t.string "email", default: "", null: false
    t.string "encrypted_password", default: "", null: false
    t.string "reset_password_token"
    t.datetime "reset_password_sent_at"
    t.datetime "remember_created_at"
    t.integer "sign_in_count", default: 0, null: false
    t.datetime "current_sign_in_at"
    t.datetime "last_sign_in_at"
    t.inet "current_sign_in_ip"
    t.inet "last_sign_in_ip"
    t.integer "role"
    t.datetime "created_at", precision: 6, null: false
    t.datetime "updated_at", precision: 6, null: false
    t.integer "sash_id"
    t.integer "level", default: 0
    t.bigint "account_id"
    t.index ["account_id"], name: "index_users_on_account_id"
    t.index ["email"], name: "index_users_on_email", unique: true
    t.index ["reset_password_token"], name: "index_users_on_reset_password_token", unique: true
  end

app / views / devise / registrations / new.html.erb

<%= simple_form_for(resource, as: resource_name, url: registration_path(resource_name)) do |f| %>
  <%= f.error_notification %>

  <div class="form-inputs">
    <%= f.input :email,
                required: true,
                autofocus: true,
                input_html: { autocomplete: "email" }%>

    <%=  f.simple_fields_for :accounts do |a| %>
      <%= a.input :name %>
    <% end %>

    <%= f.input :password,
                required: true,
                hint: ("#{@minimum_password_length} characters minimum" if @minimum_password_length),
                input_html: { autocomplete: "new-password" } %>
    <%= f.input :password_confirmation,
                required: true,
                input_html: { autocomplete: "new-password" } %>
  </div>

  <div class="form-actions">
    <%= f.button :submit, "Sign up" %>
  </div>
<% end %>

<%= render "devise/shared/links" %>

您会建议联接表account_users account：references user：references ...还是有一种简单的方法？

我考虑过一个Admin Devise模型，但这使登录有点麻烦。

Answer 1

将Account作为父项，将User作为子级，您会做得更好：

Account has_many Users

因此，您可以做的是在User模型中创建一个回调以检查是否存在帐户，如果该帐户为空，则创建一个。

before_validation :create_account_if_blank


def create_account_if_blank
    if self.account.blank?
        ApplicationRecord.transaction do
            account = Account.create!(name: self.full_name)
            some_other_thing = Something.create!(name: 'test')
        end
     end
end

然后，当您从“管理员”帐户创建另一个用户时，只需从控制器中设置当前帐户即可。

您甚至可以执行以下操作：

current_account.users.create(your parameters here)

在您的应用程序控制器中放置current_account函数。

current_account函数如下所示：

def current_account
  return current_user.account
end

Answer 2

我认为您可以使用STI来使Owner类和Employee类都从User和角色继承为Inheritance_column，然后可以在角色和帐户之间建立多态关系

class Employee < User
  has_one :account, as: :accountable 
end
class Owner < User
 has_one :account, as: :accountable 
end
# do the same with the other roles it gives you more flexibility to have different behaviour for every role than using only User class 
Account < ApplicationRecord
 belongs_to :accountable, polymorphic: true
end

设计用户拥有帐户，帐户有许多用户

2 个答案: