接受具有属于多态性的嵌套属性

我还遇到了“ArgumentError:Cannot build association model_name”的问题。你想建立一个多态的一对一关联吗?”

elsif !reject_new_record?(association_name, attributes)
  method = "build_#{association_name}"
  if respond_to?(method)
    send(method, attributes.except(*UNASSIGNABLE_KEYS))
  else
    raise ArgumentError, "Cannot build association #{association_name}. Are you trying to build a polymorphic one-to-one association?"
  end
end

那么实际上我们需要在这里做什么呢?只是为了在我们的模型中创建构建关联名。我在下面做了一个很好的例子:

class Job <ActiveRecord::Base
  CLIENT_TYPES = %w(Contact)

  attr_accessible :client_type, :client_attributes

  belongs_to :client, :polymorphic => :true

  accepts_nested_attributes_for :client

  protected

  def build_client(params, assignment_options)
    raise "Unknown client_type: #{client_type}" unless CLIENT_TYPES.include?(client_type)
    self.client = client_type.constantize.new(params)
  end
end

最后 让它与Rails 4.x一起工作。这是基于Dmitry/ScotterC的答案,所以对他们来说是+1。

第一步。 首先,这里是具有多态关联的完整模型:

# app/models/polymorph.rb
class Polymorph < ActiveRecord::Base
  belongs_to :associable, polymorphic: true

  accepts_nested_attributes_for :associable

  def build_associable(params)
    self.associable = associable_type.constantize.new(params)
  end
end

# For the sake of example:
# app/models/chicken.rb
class Chicken < ActiveRecord::Base
  has_many: :polymorphs, as: :associable
end

是的,那不是什么新鲜事。但是你可能会想,在哪里 polymorph_type 它的价值是如何设定的?它是底层数据库记录的一部分,因为多态关联添加了 <association_name>_id 和 <association_name>_type 表的列。照现在的情况,当 build_associable 执行 _type 的值为 nil .

第二步。传入并接受子类型

child_type 以及典型的表单数据,并且控制器必须允许它进行强参数检查。

# app/views/polymorph/_form.html.erb
<%= form_for(@polymorph) do |form| %>
  # Pass in the child_type - This one has been turned into a chicken!
  <%= form.hidden_field(:polymorph_type, value: 'Chicken' %>
  ...
  # Form values for Chicken
  <%= form.fields_for(:chicken) do |chicken_form| %>
    <%= chicken_form.text_field(:hunger_level) %>
    <%= chicken_form.text_field(:poop_level) %>
    ...etc...
  <% end %>
<% end %>

# app/controllers/polymorph_controllers.erb
...
private
  def polymorph_params
    params.require(:polymorph).permit(:id, :polymorph_id, :polymorph_type)
  end

希望这能帮助其他人。(你为什么需要多态鸡?)

上面的答案很好,但不适用于所示的设置。它启发了我,我创造了一个可行的解决方案:

用于创建和更新

class Job <ActiveRecord::Base
  belongs_to :client, :polymorphic=>:true
  attr_accessible :client_attributes
  accepts_nested_attributes_for :client

  def attributes=(attributes = {})
    self.client_type = attributes[:client_type]
    super
  end

  def client_attributes=(attributes)
    some_client = self.client_type.constantize.find_or_initilize_by_id(self.client_id)
    some_client.attributes = attributes
    self.client = some_client
  end
end

刚刚发现rails不支持这种行为,所以我提出了以下解决方法:

class Job <ActiveRecord::Base
  belongs_to :client, :polymorphic=>:true, :autosave=>true
  accepts_nested_attributes_for :client

  def attributes=(attributes = {})
    self.client_type = attributes[:client_type]
    super
  end

  def client_attributes=(attributes)
    self.client = type.constantize.find_or_initialize_by_id(attributes.delete(:client_id)) if client_type.valid?
  end
end

这样我就可以这样设置表单:

<%= f.select :client_type %>
<%= f.fields_for :client do |client|%>
  <%= client.text_field :name %>
<% end %>