Question

我正在尝试实现典型的gof复合模式：

我在以后查询它时有点迷失。例如，在没有任何祖先的情况下查询所有复合材料会有一个很好的方法吗？

我最初的想法是使用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

那会有用吗？我将如何构建这样的列表（在稍后的视图中f.ex。）？：

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

在查询时我只是有点迷失。这个问题有最佳实践吗？

Answer 1

实际解决方案之前有几个方面：

图表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，您的示例显示它是多对多的。
在主要使用单一模型的两种情况下都可以解决。

<强>许多对多

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

移植

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

<强>一到多

在这种情况下更简单。在组件模型中使用Ancestry（https://github.com/stefankroes/ancestry）。它将提供所需的所有操作。或者，可以使用acts_as_tree代替Ancestry。

如果您需要示例代码，请告诉我。

在Ruby on Rails中实现复合模式

1 个答案: