为什么在Ruby on Rails 3中返回Nil ?

Question

我运行了这个命令:

user1 = User.create(:email => '[e-mail]', :password => 'test12123', :username => 'test12', :first_name => 'Test', :last_name => 'User', :plan_id => '4')

并得到这个错误:

NoMethodError: undefined method `trial_duration' for nil:NilClass

这是我的用户模型:

# == Schema Information
# Schema version: 20110412170916
#
# Table name: users
#
#  id                   :integer         not null, primary key
#  email                :string(255)
#  encrypted_password   :string(128)
#  password_salt        :string(255)
#  reset_password_token :string(255)
#  remember_token       :string(255)
#  remember_created_at  :datetime
#  sign_in_count        :integer
#  current_sign_in_at   :datetime
#  last_sign_in_at      :datetime
#  current_sign_in_ip   :string(255)
#  last_sign_in_ip      :string(255)
#  username             :string(255)
#  first_name           :string(255)
#  last_name            :string(255)
#  created_at           :datetime
#  updated_at           :datetime
#  invitation_token     :string(60)
#  invitation_sent_at   :datetime
#  plan_id              :integer
#  current_state        :string(255)
#  confirmation_token   :string(255)
#  confirmed_at         :datetime
#  confirmation_sent_at :datetime
#  space_used           :integer         default(0), not null
#  failed_attempts      :integer         default(0)
#  unlock_token         :string(255)
#  locked_at            :datetime
#  trial_end_date       :date
#  active_subscription  :boolean
#
belongs_to :plan
before_create :set_trial_end
def set_trial_end
     plan = self.plan
     end_of_trial = self.created_at + plan.trial_duration.days
     self.trial_end_date = end_of_trial
end

这是我的Plan模型:

# == Schema Information
# Schema version: 20110412101615
#
# Table name: plans
#
#  id                  :integer         not null, primary key
#  name                :string(255)
#  storage             :float
#  num_of_projects     :integer
#  num_of_clients      :integer
#  cached_slug         :string(255)
#  created_at          :datetime
#  updated_at          :datetime
#  amount              :integer
#  trial_duration      :integer
#  trial_duration_unit :string(255)
#  currency            :string(255)
#  billing_cycle       :integer
#  billing_cycle_unit  :string(255)
#
class Plan < ActiveRecord::Base
  has_many  :users
    has_many    :subscriptions
    has_friendly_id :name, :use_slug => true
end

想法?

Answer 1

确保:plan_id在您的User模型的attr_accessible列表中，或者不在您的attr_protected列表中。这将防止在该用户中分配plan_id。创建声明。

Answer 2

试试这个:

plan = Plan.find(4)
user1 = User.create(:email => '[e-mail]', :password => 'test12123', :username => 'test12', :first_name => 'Test', :last_name => 'User', :plan => plan)

由于没有找到相关的"plan"对象而失败

Answer 3

您是否尝试过显式设置关联?和Simon说的差不多，但是像:

plan = Plan.find(4)
user = [code to create the user, but without the "plan" or "plan_id" attribute]
user.plan = plan
user.save

或者，在您的代码中，尝试使用:plan_id => 4而不是'4'。

Answer 4

在before_create :set_trial_end中，计划尚未提取。

我认为您需要在def set_trial_end中手动获取计划

def set_trial_end
  plan = Plan.find(plan_id)
  end_of_trial = self.created_at + plan.trial_duration.days
  self.trial_end_date = end_of_trial
end

您可能需要在创建时设置一个名为plan_id的虚拟属性。

Answer 5

也许你必须在attr_accessible中添加trial_duration。

试试这个:

class Plan < ActiveRecord::Base
has_many  :users
has_many    :subscriptions
has_friendly_id :name, :use_slug => true
attr_accessible :trial_duration
end