Question

我正在使用单表继承来管理不同类型的项目。我决定存储与每个项目类型相关的一些信息。所以我用“model_type”字段作为主键创建了新表“project_types”。主键值是“项目”表的“类型”字段的值。问题：当我尝试与Project对象ProjectTypes对象关联时，它总是返回null。

>> p = Project.find(:first)
=> #<SiteDesign id: 1, type: "SiteDesign", name: "1", description: "dddd", concept: "d", client_id: 40, created_at: "2009-10-15 08:17:45", updated_at: "2009-10-15 08:17:45">
>> p.project_type
=> nil

获取与ProjectTypes项目相关的项目是可以的。有没有办法让它正常工作？

型号：

class Project < ActiveRecord::Base
    belongs_to :project_type, :class_name => "ProjectTypes", :foreign_key => "model_name"
end

class SiteDesign < Project
end

class TechDesign < Project
end

class ProjectTypes < ActiveRecord::Base
  self.primary_key = "model_name"
  has_many :projects, :class_name => "Project", :foreign_key => "type"
end

迁移：

class CreateProjectTypes < ActiveRecord::Migration
  def self.up
    create_table :project_types, :id => false  do |t|
      t.string :model_name , :null => false
      t.string :name, :null => false
      t.text :description

      t.timestamps
    end

    add_index :project_types, :model_name, :unique => true


    #all project types that are used.
    models_names = {"SiteDesign" => "Site design",
      "TechDesign" => "Tech design"}

    #key for model_name and value for name
    models_names.each do |key,value|
      p = ProjectTypes.new();
      p.model_name = key
      p.name = value
      p.save
    end

  end

  def self.down
    drop_table :project_types
  end
end

class CreateProjects < ActiveRecord::Migration
  def self.up
    create_table :projects do |t|
      t.string :type
      t.string :name
      t.text :description
      t.text :concept
      t.integer :client_id

      t.timestamps
    end
  end

  def self.down
    drop_table :projects
  end
end

Answer 1

你遇到问题并不奇怪。通过从纯粹的STI系统转移到当前的系统，你可以通过将一部分与另一部分混合来破坏你正在使用的模式。

我个人会这样做：

class Project < ActiveRecord::Base
    attr_readonly(:project_type)
    belongs_to :project_type
    before_create :set_project_type

    def set_project_type()
        project_type = ProjectType.find_by_model_name(this.class)
    end
end

class SiteProject < Project
end

class TechProject < Project
end

class ProjectType < ActiveRecord::Base
    has_many :projects
end

迁移：

class CreateProjectTypes < ActiveRecord::Migration
  def self.up
    create_table :project_types  do |t|
      t.string :model_name , :null => false
      t.string :name, :null => false
      t.text :description

      t.timestamps
    end

    add_index :project_types, :model_name, :unique => true


    #all project types that are used.
    models_names = {"SiteDesign" => "Site design",
      "TechDesign" => "Tech design"}

    #key for model_name and value for name
    models_names.each do |key,value|
      p = ProjectTypes.new();
      p.model_name = key
      p.name = value
      p.save
    end

  end

  def self.down
    drop_table :project_types
  end
end

class CreateProjects < ActiveRecord::Migration
  def self.up
    create_table :projects do |t|
      t.string :type
      t.references :project_type, :null => false
      t.text :description
      t.text :concept
      t.integer :client_id

      t.timestamps
    end
  end

  def self.down
    drop_table :projects
  end
end

它只是清理事物，它也有助于澄清你正在做的事情。您的“ProjectType”表纯粹用于额外数据，您的继承树仍然存在。我还投入了一些检查，以确保您的项目类型始终设置（并且正确地，基于模型名称），并通过使该属性为只读来阻止您更改项目类型。

STI和has_many将“type”列关联为Key

1 个答案: