将Rails STI类型参数转换为实际的类常量？

Question

我是Rails新手。我使用以下class User及其class Teacher < User子类创建STI。当我填写用户表格时，它引发了错误。

/ users处的NameError

  

未初始化的常量User :: teacher

app / controllers / users_controller.rb，第29行

   24     end
   25   
   26     # POST /users
   27     def create 
   28       byebug 
>  29       @user = User.new(user_params) 
   30       authorize @user    
   31       if @user.save
   32         render json: @user, status: :created, location: @user, serializer: UserSerializer
   33       else
   34         render json: @user.errors, status: :unprocessable_entity

应用回溯

• app / controllers / users_controller.rb：29：in'create'

完整追溯

activerecord (6.0.1) lib/active_record/inheritance.rb:206:in    `compute_type'
activerecord (6.0.1) lib/active_record/inheritance.rb:233:in `find_sti_class'
activerecord (6.0.1) lib/active_record/inheritance.rb:263:in `subclass_from_attributes'
activerecord (6.0.1) lib/active_record/inheritance.rb:56:in `new'
app/controllers/users_controller.rb:29:in `create'
actionpack (6.0.1) lib/action_controller/metal/basic_implicit_render.rb:6:in    `send_action'
actionpack (6.0.1) lib/abstract_controller/base.rb:196:in `process_action'
actionpack (6.0.1) lib/action_controller/metal/rendering.rb:30:in `process_action'
actionpack (6.0.1) lib/abstract_controller/callbacks.rb:42:in `block in process_action'
activesupport (6.0.1) lib/active_support/callbacks.rb:135:in `run_callbacks'
actionpack (6.0.1) lib/abstract_controller/callbacks.rb:41:in `process_action'
actionpack (6.0.1) lib/action_controller/metal/rescue.rb:22:in `process_action'
actionpack (6.0.1) lib/action_controller/metal/instrumentation.rb:33:in `block in    process_action'
activesupport (6.0.1) lib/active_support/notifications.rb:180:in `block in instrument'
activesupport (6.0.1) lib/active_support/notifications/instrumenter.rb:24:in `instrument'
activesupport (6.0.1) lib/active_support/notifications.rb:180:in `instrument'
actionpack (6.0.1) lib/action_controller/metal/instrumentation.rb:32:in `process_action'
actionpack (6.0.1) lib/action_controller/metal/params_wrapper.rb:245:in `process_action'
activerecord (6.0.1) lib/active_record/railties/controller_runtime.rb:27:in    `process_action'
actionpack (6.0.1) lib/abstract_controller/base.rb:136:in `process'
actionpack (6.0.1) lib/action_controller/metal.rb:191:in `dispatch'
actionpack (6.0.1) lib/action_controller/metal.rb:252:in `dispatch'
actionpack (6.0.1) lib/action_dispatch/routing/route_set.rb:51:in `dispatch'
actionpack (6.0.1) lib/action_dispatch/routing/route_set.rb:33:in `serve'
actionpack (6.0.1) lib/action_dispatch/journey/router.rb:49:in `block in serve'
actionpack (6.0.1) lib/action_dispatch/journey/router.rb:32:in `serve'
actionpack (6.0.1) lib/action_dispatch/routing/route_set.rb:837:in `call'

我想我缺少将字符串参数常量化为实际类常量的部分。怎么做？

请指导。谢谢

Answer 1

通过查看uninitialized constant User::teacher，确保您在数据库中也有类型，好像您在某个地方使用的是较低版本。应该是User::Teacher

Answer 2

如Paula Fidalgo所说，请确保您在用户模型架构中将column类型作为字符串输入

之后，您可以在用户模型中创建新条目，如下所示：

Teacher.create(user_params)

User.create(type: 'Teacher')

都一样。