Question

我不知道出了什么问题，但我无法使用belongs_to：class_name选项。有人可以启发我吗？非常感谢！

以下是我的代码中的剪辑。

class CreateUsers < ActiveRecord::Migration
    def self.up
        create_table :users do |t|
            t.text :name
        end
    end

    def self.down
        drop_table :users
    end
end

#####################################################

class CreateBooks < ActiveRecord::Migration
    def self.up
        create_table :books do |t|
            t.text :title
            t.integer :author_id, :null => false
        end
    end

    def self.down
        drop_table :books
    end
end

#####################################################

class User < ActiveRecord::Base
    has_many: books
end

#####################################################

class Book < ActiveRecord::Base
    belongs_to :author, :class_name => 'User', :validate => true
end

#####################################################

class BooksController < ApplicationController
    def create
        user = User.new({:name => 'John Woo'})
        user.save
        @failed_book = Book.new({:title => 'Failed!', :author => @user})
        @failed_book.save # missing author_id
        @success_book = Book.new({:title => 'Nice day', :author_id => @user.id})
        @success_book.save # no error!
    end
end

环境：

ruby 1.9.1-p387 Rails 2.3.5

Answer 1

class User < ActiveRecord::Base
  has_many :books, :foreign_key => 'author_id'
end

class Book < ActiveRecord::Base
  belongs_to :author, :class_name => 'User', :foreign_key => 'author_id', :validate => true
end

最好的办法是更改迁移并将author_id更改为user_id。然后，您可以删除:foreign_key选项。

Answer 2

应该是

belongs_to :user, :foreign_key => 'author_id'

如果您的外键是作者ID。由于您实际拥有User类，因此您的Book必须为belongs_to：user。

Answer 3

我这样做：

迁移 -

class AddAuthorToPosts < ActiveRecord::Migration
  def change
    add_reference :posts, :author, index: true
    add_foreign_key :posts, :users, column: :author_id
  end
end

课程帖子

  belongs_to :author, class_name: "User"

Answer 4

迁移

t.belongs_to :author, foreign_key: { to_table: :users }

在轨道5上工作

belongs_to with：class_name选项失败

4 个答案: