我正在尝试实现典型的gof复合模式:
example class diagram
稍后再查询时,我有点迷失。
例如,有没有任何祖先的好方法来查询所有Composites?
我最初的想法是用ActiveRecord创建类​​似的东西

class Component < ActiveRecord::Base
  belongs_to :childrenable, :polymorphic => true
  has_and_belongs_to_many: composites
end

class Leaf < ActiveRecord::Base
  has_many: components, :as => :childrenable
end

class Composite < ActiveRecord::Base
  has_many: components, :as => :childrenable
  has_and_belongs_to_many :components
end

那行得通吗?我将如何建立这样的列表(在稍后的f.ex.中的视图中)?:
CompositeA
  ->Item
  ->CompositeB
    ->ItemA
  ->CompositeC
    ->ItemA
    ->ItemB

当涉及到查询时,我只是有点迷路。是否有针对此问题的最佳实践?

最佳答案

实际解决方案之前有两个方面:


该图和您的示例在非常关键的方面有所不同。该图表明容器与子代之间的关系是一对多的。但是,您的示例表明它是多对多的。
在两种情况下都可以使用一个模型来解决。


多对多

可以使用与自己的多对多关系来解决。

模型

class Component < ActiveRecord::Base
  # Add as many attributes you need
  attr_accessible :name

  has_and_belongs_to_many :children,
    :class_name => "Component",
    :join_table => "children_containers",
    :foreign_key => "container_id",
    :association_foreign_key => "child_id"

  has_and_belongs_to_many :containers,
    :class_name => "Component",
    :join_table => "children_containers",
    :foreign_key => "child_id",
    :association_foreign_key => "container_id"

  # All Components that do not belong to any container
  scope :roots, -> {where("not exists (select * from children_containers where child_id=components.id)")}

  # All Components that have no children
  scope :leaves, -> {where("not exists (select * from children_containers where container_id=components.id)")}

  # Is this Component at root level
  def root?
    self.containers.empty?
  end

  # Is this Component at leaf level
  def leaf?
    self.children.empty?
  end

  # Notice the recursive call to traverse the Component hierarchy
  #   Similarly, it can be written to output using nested <ul> and <li>s as well.
  def to_s(level=0)
    "#{'  ' * level}#{name}\n" + children.map {|c| c.to_s(level + 1)}.join
  end
end


移民

class CreateComponents < ActiveRecord::Migration
  def change
    create_table :components do |t|
      t.string :name

      t.timestamps
    end

    create_table :children_containers, :id => false do |t|
        t.references :child
        t.references :container
    end

    add_index :children_containers, :child_id
    add_index :children_containers, [:container_id, :child_id], :unique => true
  end
end


样例代码

["R1", "R2", "L1", "L2", "C1", "C2", "C3"].each {|n| Component.create(:name => n)}

[
    ["R1", "C1"],
    ["R2", "C2"],
    ["R1", "C3"],
    ["R2", "C3"],
    ["C1", "L1"],
    ["C2", "L2"],
    ["C3", "L1"],
    ["C3", "L2"]
].each {|pair| p,c=pair; Component.find_by_name(p).children << Component.find_by_name(c)}

puts Component.roots.map(&:name).to_s
# ["R1", "R2"]

puts Component.leaves.map(&:name).to_s
# ["L1", "L2"]

puts Component.find_by_name("R1").to_s
# R1
#   C1
#     L1
#   C3
#     L1
#     L2


一对多

在这种情况下,它要简单得多。在组件模型中使用祖先(https://github.com/stefankroes/ancestry)。它将提供所有需要的操作。或者,可以使用acts_as_tree代替Ancestry。

让我知道您是否需要示例代码。

09-18 03:51