4

我正在尝试实现典型的 gof 复合模式：

示例类图

稍后查询它时，我有点迷茫。例如，是否有一种很好的方法来查询没有任何祖先的所有复合材料？

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？我将如何构建这样的列表（在 f.ex. 的稍后视图中）？：

CompositeA  
  ->Item
  ->CompositeB
    ->ItemA
  ->CompositeC
    ->ItemA
    ->ItemB

当涉及到查询时，我有点迷茫。这个问题有什么最佳实践吗？

4

1 回答 1

7

在实际解决方案之前有几个方面：

  • 该图和您的示例在一个非常关键的方面有所不同。该图表明 Container 和 Children 之间的关系是一对多的。但是，您的示例表明它是多对多的。
  • 它可以在这两种情况下主要使用一个模型来解决。

多对多

它可以使用与自身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

示例代码

["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

在这种情况下要简单得多。在组件模型中使用 Ancestry ( 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 代替祖先。

如果您需要此示例代码，请告诉我。

于 2013-07-19T22:59:34.397 回答