我已经设置了一个Rails项目来使用Single Table Inheritance,因为我有User
s - Sender
s和Receiver
s两种类型。 Sender
s有public_key
财产和Receiver
s有phone_number
财产。他们通过User
共享name
,email
和password
属性。Rails - 如何从单表继承创建正确类型的对象?
我的问题是,在User
控制器的create
功能,我试图创建一个类型或其他 - 无论是Sender
或Receiver
- 基于单选按钮的我的注册表格上的数值。
这里的设置:
用户模型
class User < ActiveRecord::Base
validates :name, :presence => true, :length => { maximum: 50 }
validates :password, presence: true
self.inheritance_column = :user_type
# We will need a way to know which types with subclass the User model
def self.user_types
%w(Sender Receiver)
end
end
class Sender < User; end
class Receiver < User; end
SENDER模型
class Sender < User
validates :public_key, :presence => true
end
接收器型号
class Receiver < User
validates :phone_number, :presence => true, :length => 10
end
用户控制器
class UsersController < ApplicationController
# before_action :set_user, only: [:show, :edit, :update, :destroy]
before_action :set_user_type
def index
@users = user_type_class.all
end
def show
@user = User.find(params[:id])
end
def new
@user = User.new
end
def create
if(user_type.eql? "receiver")
@user = Receiver.new(user_params)
else
@user = Sender.new(user_params)
end
...
private
# allow views to access user_type
def set_user_type
@user_type = user_type
end
def user_type
User.user_types.include?(params[:type]) ? params[:type] : "User"
end
def user_type_class
user_type.constantize
end
# Use callbacks to share common setup or constraints between actions.
def set_user
@user = User.find(params[:id])
end
def user_params
params.require(:user).permit(:name, :email, :password, :confirmation_password, :user_type)
end
end
New User表单
<%= form_for(@user) do |f| %>
...
<div class="user-type">
<%= f.label :user_type, class: 'form-control' %>
<%= radio_button_tag(:user_type, "sender") %>
<%= label_tag(:user_type_sender, "I am a Sender") %>
<%= radio_button_tag(:user_type, "receiver") %>
<%= label_tag(:user_type_receiver, "I am a Receiver") %>
</div>
<div class="field">
<%= f.label :name %><br>
<%= f.text_field :name %>
</div>
...
<% end %>
在User
控制器的create
方法,我试图与这if
说法正确类型的User
:
if(user_type.eql? "receiver")
@user = Receiver.new(user_params)
else
@user = Sender.new(user_params)
基于在此处User
表格中记录的值:
<%= radio_button_tag(:user_type, "sender") %>
<%= label_tag(:user_type_sender, "I am a Sender") %>
<%= radio_button_tag(:user_type, "receiver") %>
<%= label_tag(:user_type_receiver, "I am a Receiver") %>
但是,我总是从else
语句结束Sender
类型对象。我在想这意味着我的if
声明有错,if(user_type.eql? "receiver")
;然而,我无法弄清楚什么。
想法?
你说得对,那是正确的 –