Question

我正在为我们的铁路教堂写一个导师计划（我仍然很喜欢铁路）..

我需要对此进行建模..

contact
has_one :father, :class_name => "Contact"
has_one :mother, :class_name => "Contact"
has_many :children, :class_name => "Contact"
has_many :siblings, :through <Mother and Father>, :source => :children

所以基本上一个对象“兄弟姐妹”需要映射父亲和母亲的所有孩子，不包括对象本身。

这可能吗？

由于

丹尼尔

Answer 1

如果看似简单的问题可以得到复杂的答案，这很有趣。在这种情况下，实现反身父/子关系相当简单，但添加父/母和兄弟关系会产生一些曲折。

首先，我们创建表来保存父子关系。关系有两个外键，都指向Contact：

create_table :contacts do |t|
  t.string :name
end

create_table :relationships do |t|
  t.integer :contact_id
  t.integer :relation_id
  t.string :relation_type
end

在关系模型中，我们将父亲和母亲指回联系人：

class Relationship < ActiveRecord::Base
  belongs_to :contact
  belongs_to :father, :foreign_key => :relation_id, :class_name => "Contact",
  :conditions => { :relationships => { :relation_type => 'father'}}
  belongs_to :mother, :foreign_key => :relation_id, :class_name => "Contact",
  :conditions => { :relationships => { :relation_type => 'mother'}}
end

并在Contact中定义反向关联：

class Contact < ActiveRecord::Base
  has_many :relationships, :dependent => :destroy
  has_one :father, :through => :relationships
  has_one :mother, :through => :relationships
end

现在可以创建一个关系：

@bart = Contact.create(:name=>"Bart")
@homer = Contact.create(:name=>"Homer")
@bart.relationships.build(:relation_type=>"father",:father=>@homer)
@bart.save!
@bart.father.should == @homer

这不是很好，我们真正想要的是在一次通话中建立关系：

class Contact < ActiveRecord::Base
  def build_father(father)
    relationships.build(:father=>father,:relation_type=>'father')
  end
end

所以我们可以这样做：

@bart.build_father(@homer)
@bart.save!

要查找联系人的子项，请向Contact和（为方便起见）实例方法添加范围：

scope :children, lambda { |contact| joins(:relationships).\
  where(:relationships => { :relation_type => ['father','mother']}) }

def children
  self.class.children(self)
end

Contact.children(@homer) # => [Contact name: "Bart")]
@homer.children # => [Contact name: "Bart")]

兄弟姐妹是棘手的部分。我们可以利用Contact.children方法并操纵结果：

def siblings
  ((self.father ? self.father.children : []) +
   (self.mother ? self.mother.children : [])
   ).uniq - [self]
end

这不是最优的，因为father.children和mother.children将重叠（因此需要uniq），并且可以通过计算必要的SQL（左侧作为练习:)来更有效地完成），但要记住self.father.children和self.mother.children在半兄弟姐妹（同一个父亲，不同的母亲）的情况下不会重叠，并且联系人可能没有父亲或母亲。< / p>

以下是完整的模型和一些规格：

# app/models/contact.rb
class Contact < ActiveRecord::Base
  has_many :relationships, :dependent => :destroy
  has_one :father, :through => :relationships
  has_one :mother, :through => :relationships

  scope :children, lambda { |contact| joins(:relationships).\
    where(:relationships => { :relation_type => ['father','mother']}) }

  def build_father(father)
    # TODO figure out how to get ActiveRecord to create this method for us
    # TODO failing that, figure out how to build father without passing in relation_type
    relationships.build(:father=>father,:relation_type=>'father')
  end

  def build_mother(mother)
    relationships.build(:mother=>mother,:relation_type=>'mother')
  end

  def children
    self.class.children(self)
  end

  def siblings
    ((self.father ? self.father.children : []) +
     (self.mother ? self.mother.children : [])
     ).uniq - [self]
  end
end

# app/models/relationship.rb
class Relationship < ActiveRecord::Base
  belongs_to :contact
  belongs_to :father, :foreign_key => :relation_id, :class_name => "Contact",
  :conditions => { :relationships => { :relation_type => 'father'}}
  belongs_to :mother, :foreign_key => :relation_id, :class_name => "Contact",
  :conditions => { :relationships => { :relation_type => 'mother'}}
end

# spec/models/contact.rb
require 'spec_helper'

describe Contact do
  before(:each) do
    @bart = Contact.create(:name=>"Bart")
    @homer = Contact.create(:name=>"Homer")
    @marge = Contact.create(:name=>"Marge")
    @lisa = Contact.create(:name=>"Lisa")
  end

  it "has a father" do
    @bart.relationships.build(:relation_type=>"father",:father=>@homer)
    @bart.save!
    @bart.father.should == @homer
    @bart.mother.should be_nil
  end

  it "can build_father" do
    @bart.build_father(@homer)
    @bart.save!
    @bart.father.should == @homer
  end

  it "has a mother" do
    @bart.relationships.build(:relation_type=>"mother",:father=>@marge)
    @bart.save!
    @bart.mother.should == @marge
    @bart.father.should be_nil
  end

  it "can build_mother" do
    @bart.build_mother(@marge)
    @bart.save!
    @bart.mother.should == @marge
  end

  it "has children" do
    @bart.build_father(@homer)
    @bart.build_mother(@marge)
    @bart.save!
    Contact.children(@homer).should include(@bart)
    Contact.children(@marge).should include(@bart)
    @homer.children.should include(@bart)
    @marge.children.should include(@bart)
  end

  it "has siblings" do
    @bart.build_father(@homer)
    @bart.build_mother(@marge)
    @bart.save!
    @lisa.build_father(@homer)
    @lisa.build_mother(@marge)
    @lisa.save!
    @bart.siblings.should == [@lisa]
    @lisa.siblings.should == [@bart]
    @bart.siblings.should_not include(@bart)
    @lisa.siblings.should_not include(@lisa)
  end

  it "doesn't choke on nil father/mother" do
    @bart.siblings.should be_empty
  end
end

Answer 2

我完全同意zetetic。这个问题看起来比答案简单得多，我们几乎无能为力。我会添加我的20c 表：

    create_table :contacts do |t|
      t.string :name
      t.string :gender
    end
    create_table :relations, :id => false do |t|
      t.integer :parent_id
      t.integer :child_id
    end

表关系没有相应的模型。

class Contact < ActiveRecord::Base
  has_and_belongs_to_many :parents,
    :class_name => 'Contact',
    :join_table => 'relations',
    :foreign_key => 'child_id',
    :association_foreign_key => 'parent_id'

  has_and_belongs_to_many :children,
    :class_name => 'Contact',
    :join_table => 'relations',
    :foreign_key => 'parent_id',
    :association_foreign_key => 'child_id'

  def siblings
    result = self.parents.reduce [] {|children, p| children.concat  p.children}
    result.uniq.reject {|c| c == self}
  end

  def father
    parents.where(:gender => 'm').first
  end

  def mother
    parents.where(:gender => 'f').first
  end
end

现在我们有定期的Rails协议。所以我们可以

alice.parents << bob
alice.save

bob.chidren << cindy
bob.save

alice.parents.create(Contact.create(:name => 'Teresa', :gender => 'f')

以及所有类似的东西。

Answer 3

  has_and_belongs_to_many :parents,
    :class_name => 'Contact',
    :join_table => 'relations',
    :foreign_key => 'child_id',
    :association_foreign_key => 'parent_id',
    :delete_sql = 'DELETE FROM relations WHERE child_id = #{id}'

  has_and_belongs_to_many :children,
    :class_name => 'Contact',
    :join_table => 'relations',
    :foreign_key => 'parent_id',
    :association_foreign_key => 'child_id',
    :delete_sql = 'DELETE FROM relations WHERE parent_id = #{id}'

我使用了这个例子但是必须添加：delete_sql来清理关系记录。起初我在字符串周围使用双引号但发现导致错误。切换到单引号有效。

has_many：通过多个has_one关系？

3 个答案: