Question

我有以下协会

class Training < ApplicationRecord
  has_many :attendances
  has_many :attendees, through: :attendances
end

class Attendance < ApplicationRecord
  belongs_to :training
  belongs_to :attendee, class_name: 'Employee'

出勤表格包含attendee_id和training_id。

现在，如何使用FactoryGirl创建有效的Attendance

目前，我有以下代码

FactoryGirl.define do
  factory :attendance do
    training
    attendee
  end
end

FactoryGirl.define do
  factory :employee, aliases: [:attendee] do
    sequence(:full_name) { |n| "John Doe#{n}" }
    department
  end
end

但是我得到了

  NoMethodError:
       undefined method `employee=' for #<Attendance:0x007f83b163b8e8>

我也试过

FactoryGirl.define do
  factory :attendance do
    training
    association :attendee, factory: :employee
  end
end

结果相同。

感谢您的帮助（或SO上不允许礼貌）。

Answer 1

您可能已经知道FactoryGirl使用该符号来推断该类是什么，但是当您为同一模型创建另一个具有不同符号的工厂时，您需要告诉FactoryGirl要使用的类是什么：

FactoryGirl.define do
  factory :attendance do
    training = { FactoryGirl.create(:training) }
    attendee = { FactoryGirl.create(:employee) }
  end
end

FactoryGirl.define do
  factory :employee, class: Attendee do
    sequence(:full_name) { |n| "John Doe#{n}" }
    department
  end
end

或者可以手动分配关系（例如，如果您此时不希望将员工实例保存到数据库）：

FactoryGirl.build(:attendance, attendee: FactoryGirl.build(:employee))

FactoryGirl与不同名称的关联

1 个答案: