Question

我期待混合mongoid和simple_form的一些麻烦（我认为 - 同样的问题对于其他表单构建者来说是实际的）。我有一个关系的小形式，如

f.input :author, collection: User.all, as: :select

当我提交没有选择作者的表格时，我会看到像

这样的例外情况

NoMethodError in ExamplesController#update
undefined method `id' for "":String

太伤心了:(正如我所见 - simple_form提交“”（空字符串）但不是nil到控制器。当然 - 我可以从表单构建器验证每个参数，但我不确定它是不是很好的解决方案。可以你推荐我什么？

UPD （模型结构）：

user.rb

class User
  include Mongoid::Document
  include Mongoid::Timestamps

  has_many :examples, :inverse_of => :author
  has_many :examples, :inverse_of => :responsible_person
end

example.rb

class Example
  include Mongoid::Document
  include Mongoid::Timestamps
  include Mongoid::MultiParameterAttributes

  field :title, type: String
  field :description, type: String

  belongs_to :author, :class_name => "User"
  belongs_to :responsible_person, :class_name => "User"

  validates_presence_of :title, :description, :author

  attr_accessible :title, :description, :author, :responsible_person

end

Answer 1

我在表单上直接使用author_id和responsible_person_id解决了这个问题，就像这样

= f.input :author_id, collection: User.all, as: :select
= f.input :responsible_person_id, collection: User.all, as: :select

而不是

= f.input :author, collection: User.all, as: :select
= f.input :responsible_person, collection: User.all, as: :select

Answer 2

试试这个

<%= f.association :author %>

mongoid，形成建设者和关系

2 个答案: