Question

我想在rails中创建自引用关系。我有一个人物模型，该人应该拥有具有相同Person对象的主人和学生。

到目前为止，我试过了：

class Person <ActiveRecord::Base
   has_many :relationships, :dependent => :destroy
   has_many :masters, :through => :relationships, :conditions => "status='master'"
   has_many :pupils, :through => :relationships, :conditions => "status='pupil'"
   has_many :inverse_relationships, :class_name => "Relationship",
      :foreign_key => "related_id"
   has_many :inverse_masters, :through => :inverse_relationships,  
      :source => :person, :conditions => "status='master'"
   has_many :inverse_pupils, :through => :inverse_relationships,  
      :source => :person, :conditions => "status='pupil'"
end

class Relationship < ActiveRecord::Base
  belongs_to :person
  belongs_to :master, :class_name => "Person", :foreign_key => 'related_id'
  belongs_to :pupil, :class_name => "Person", :foreign_key => 'related_id'
end

当我尝试选择时似乎有效：

@a = Person.find(:first)
@a.masters

但是当我尝试推送到主人时，它会保存关系，而不会将状态设置为主人。它保存了null。当我进入主人时，是否有一种简单的方法可以保存status=master;当我推入学生时，status=pupil是否可以保存？{/ p>

由于

Answer 1

简而言之，解决方案是：关联回调（更多来自关联回调部分：http://railsapi.com/doc/rails-v3.0.0/classes/ActiveRecord/Associations/ClassMethods.html）

为了更详细一些，我稍微调整了你的例子，但基本上结构是一样的，这里是代码：

class Person < ActiveRecord::Base
  has_many :relationships
  has_many :pupils, :through => :relationships, :source => :other_person, :conditions => 'relationships.type = "MasterPupil"', :after_add => Proc.new{|p,o| Relationship.update_all("type = 'MasterPupil'", ['person_id = ? AND other_person_id = ?', p.id, o.id])}
  has_many :masters, :through => :relationships, :source => :other_person, :conditions => 'relationships.type = "PupilMaster"', :after_add => Proc.new{|p,o| Relationship.update_all("type = 'PupilMaster'", ['person_id = ? AND other_person_id = ?', p.id, o.id])}
end

class Relationship < ActiveRecord::Base
  belongs_to :person
  belongs_to :other_person, :class_name => 'Person'
  before_validation :set_type

  def set_type
    self.type = 'OpenRelationship'
  end
end

class MasterPupil < Relationship
end

class PupilMaster < Relationship
end

RelationShip模型包含一个类型列，它等同于您的状态列，但如果我以后想要进行STI并声明MasterPupil / PupilMaster关系模型，则类型更好。

RelationShip还有一个set_type before_validation，它将类型设置为OpenRelationship，它应该是临时的，然后在每个关联中Person模型中定义的after_add回调将设置清楚（并设置MasterPupil或PupilMaster类型）

现在：

Loading development environment (Rails 3.0.0)
irb(main):001:0> p = Person.create
=> #<Person id: 1, created_at: "2010-09-10 23:35:37", updated_at: "2010-09-10 23:35:37">
irb(main):002:0> p.pupils
=> []
irb(main):003:0> p.masters
=> []
irb(main):004:0> p.pupils << Person.create
=> [#<Person id: 2, created_at: "2010-09-10 23:35:56", updated_at: "2010-09-10 23:35:56">]
irb(main):005:0> Relationship.all
=> [#<MasterPupil id: 1, person_id: 1, other_person_id: 2, type: "MasterPupil">]
irb(main):006:0> p.masters << Person.create
=> [#<Person id: 3, created_at: "2010-09-10 23:36:29", updated_at: "2010-09-10 23:36:29">]
irb(main):007:0> Relationship.all
=> [#<MasterPupil id: 1, person_id: 1, other_person_id: 2, type: "MasterPupil">, #<PupilMaster id: 2, person_id: 1, other_person_id: 3, type: "PupilMaster">]
irb(main):008:0> p.reload
=> #<Person id: 1, created_at: "2010-09-10 23:35:37", updated_at: "2010-09-10 23:35:37">
irb(main):009:0> p.pupils
=> [#<Person id: 2, created_at: "2010-09-10 23:35:56", updated_at: "2010-09-10 23:35:56">]
irb(main):010:0> p.masters
=> [#<Person id: 3, created_at: "2010-09-10 23:36:29", updated_at: "2010-09-10 23:36:29">]

使用polymorph多对多自引用与rails中关系的属性的关系

1 个答案: