Rails ActiveRecord 两个字段，每个字段都引用同一个表答案

【问题标题】：Rails ActiveRecord two fields each referencing the same tableRails ActiveRecord 两个字段，每个字段都引用同一个表
【发布时间】：2020-01-31 00:14:12
【问题描述】：

我有一个 Player 类，我想拥有 high_school_team 和 club_team 属性。那么我认为Player 将具有指向相应团队的high_school_team_id 和club_team_id 属性。我尝试在以下迁移中执行此操作，但它不起作用。

class CreatePlayers < ActiveRecord::Migration[6.0]
  def change
    create_table :players do |t|
      t.string :first_name
      t.string :middle_name
      t.string :last_name
      t.decimal :height
      t.decimal :weight
      t.date :birthday
      t.references :team, :high_school_team, foreign_key: true
      t.references :team, :club_team, foreign_key: true
      t.decimal :gpa
      t.string :class_year
      t.string :intended_major
      t.string :email
      t.string :phone_number
      t.text :notes

      t.timestamps
    end
  end
end

它给出了以下错误：

code/scout-db [master●] » rails db:migrate
== 20191218003854 CreatePlayers: migrating ====================================
-- create_table(:players)
rails aborted!
StandardError: An error has occurred, this and all later migrations canceled:

you can't define an already defined column 'team_id'.
/Library/Ruby/Gems/2.6.0/gems/activerecord-6.0.2.1/lib/active_record/connection_adapters/abstract/schema_definitions.rb:372:in `column'
...

HighSchoolTeam 和ClubTeam 是与Team 进行单表继承的模型。

我不明白为什么会出现错误。 docs 似乎说t.referenes 的第一个参数是table_name 和第二个ref_name。 :team 是表的名称，我希望引用为 high_school_team_id 和 club_team_id。

当我将参数的顺序切换到t.references 时，它仍然不起作用。不知何故，它给出了同样的错误：you can't define an already defined column 'team_id'.。

【问题讨论】：

标签： ruby-on-rails rails-activerecord

【解决方案1】：

看来您混淆了SchemaStatement#add_reference 和TableDefinition#references，它们的签名完全不同。

如果您想设置一个无法从列名（第一个参数）派生表的外键列，您只需传递foreign_key: { to_table: :teams}。

class CreatePlayers < ActiveRecord::Migration[6.0]
  def change
    create_table :players do |t|
      t.references :high_school_team, foreign_key: { to_table: :teams}
      t.references :club_team, foreign_key: { to_table: :teams}
    end
  end
end

t.references :high_school_team, index: true 正如其他答案所推荐的那样，并不等同。这只是向列添加索引，但没有外键约束。

然后您可以将 Player 上的关联设置为：

class Player < ApplicationRecord
  belongs_to :high_school_team, class_name: 'Team'
  belongs_to :club_team, class_name: 'Team'
end

您不能在另一端使用单个 has_many :players 关联，因为外键列可以是 players.high_school_team_id 或 players.club_team_id。

class Team
  has_many :high_school_team_players, 
     foreign_key: :high_school_team_id,
     class_name: 'Player'
  has_many :club_team_players, 
     foreign_key: :club_team_id,
     class_name: 'Player'
end

但首先真正更好的选择是设置一个连接表：

class Player
  has_many :placements
  has_one :high_school_team, 
    through: :placements,
    source: :team
    class_name: 'Team'
  has_one :club_team, 
    through: :placements,
    source: :team
    class_name: 'Team'
end

class Placement
  belongs_to :player
  belongs_to :team
end

class Team
  has_many :placements
  has_many :players, through: :placements  
end

【讨论】：

add_reference 的等效调用是 add_reference(:teams, :high_school_team, foreign_key: {to_table: :teams })

【解决方案2】：

您提到的文档谈到了需要添加对现有表的引用的情况。

用于添加对新表的引用：

t.references :team, :high_school_team, foreign_key: true

这段代码是错误的。相反，它应该是

t.references :high_school_team, foreign_key: {to_table: :teams}

to_table需要添加database referential integrity

所以你的迁移会是这样的：

class CreatePlayers < ActiveRecord::Migration[6.0]
  def change
    create_table :players do |t|
      ....
      t.references :high_school_team, foreign_key: {to_table: :teams}
      t.references :club_team, foreign_key: {to_table: :teams}
      ....
    end
  end
end

【讨论】：