Question

如果有孩子，我试图阻止记录被破坏。

class Submission < ActiveRecord::Base

has_many :quotations, :dependent => :destroy

 before_destroy :check_for_payments


  def quoted?
    quotations.any?
  end


  def has_payments?
   true if quotations.detect {|q| q.payment}
  end


  private

  def check_for_payments
    if quoted? && has_payments?
      errors[:base] << "cannot delete submission that has already been paid"
      false
    end
  end

end

class Quotation < ActiveRecord::Base

    #associations
    belongs_to :submission
        has_one :payment_notification   
        has_one :payment

         before_destroy :check_for_payments

private 

def check_for_payments
  if payment_notification || payment
    errors[:base] << "cannot delete quotation while payment exist"
    return false
  end
end
end

当我测试此代码时，before_destroy：check_for_payments会阻止删除引用记录。

然而，提交before_destroy回调中的：check_for_payments并不会阻止提交被删除。

如何通过销毁付款来停止提交？

Answer 1

在 Rails 5 中，你必须my-slideshow否则它将无效。（甚至返回throw :abort）

这样的事情应该有效：

false

Answer 2

http://api.rubyonrails.org/classes/ActiveRecord/Callbacks.html 订购回调（我改变了这个具体例子的措辞）

有时代码需要回调在特定的情况下执行   订购。例如，before_destroy回调（check_for_payments in   这个案例）应该在报价被销毁之前执行   + dependent：destroy +选项。

在这种情况下，问题是before_destroy回调是   执行时，引用不可用，因为destroy回调   先被执行。您可以使用前置选项   before_destroy回调以避免这种情况。

before_destroy :check_for_payments, prepend: true

我使用与您上述相同的模型创建了一个新应用程序，然后进行了提交测试。它非常难看，我只是在学习......

class Submission < ActiveRecord::Base

  has_many :quotations, :dependent => :destroy

  before_destroy :check_for_payments, prepend: true

  def quoted?
    quotations.any?
  end

  def has_payments?
    true if quotations.detect {|q| q.payment }
  end

  private

    def check_for_payments
      if quoted? && has_payments?
        errors[:base] << "error message"
        false
      end
    end

end

class Quotation < ActiveRecord::Base

  belongs_to :submission
  has_one :payment_notification   
  has_one :payment

  before_destroy :check_for_payments

  private 

    def check_for_payments
      if payment_notification || payment
        errors[:base] << "cannot delete quotation while payment exist"
        return false
      end
    end
end

require 'test_helper'

class SubmissionTest < ActiveSupport::TestCase


  test "can't destroy" do

    sub = Submission.new
    sub.save

    quote = Quotation.new
    quote.submission_id = sub.id
    quote.save

    pay = Payment.new
    pay.quotation_id = quote.id
    pay.save

    refute sub.destroy, "destroyed record"
  end
end

它过去了！我希望有所帮助。

Answer 3

我会尝试下面的代码：

使用了has_many：通过关联付款
通过使用any?而不使用阻止导致使用关联计数器缓存（如果已定义），或者如果已加载关联的大小而导致使用SQL COUNT（如果需要）失败，则避免对报价和付款进行不必要的记录检索。
避免列举报价
避免直接测试q.payment关联代理的真实性/存在性，这对has_xxx不起作用。如果您要测试在线使用q.payment.present?

尝试以下操作，看看你如何：

class Submission < ActiveRecord::Base

  has_many :quotations,
    inverse_of: :submission,
    dependent: :destroy

  has_many :payments,
    through: :quotations

  before_destroy :check_for_payments, prepend: true

private

  def check_for_payments
    if payments.any?
      errors[:base] << "cannot delete submission that has already been paid"
      return false
    end
  end
end

Answer 4

据我所知，当对与Submission有关联的类（dependent => :destroy）的对象调用destroy时，如果相关模型中的任何回调失败，在你的情况下Quotation，然后Submission类对象仍将被删除。

因此，为了防止这种行为，我们必须采用目前可以想到的方法：

1）不是在Quotation#check_for_payments中返回false，而是可以在Submission模型中引发异常并优雅地处理它，这将完成ROLLBACK，并且不允许任何记录被摧毁。

2）您可以检查quotations实例的任何Submission是否在payment方法本身中具有payment_notification / Submission#check_for_payments，这会阻止删除Submission对象。

Answer 5

确保quoted?和has_payments?不会返回false。

有关调试，请尝试此

def check_for_payments
    raise "Debugging #{quoted?} #{has_payments?}" # Make sure both are true
    if quoted? && has_payments?
      errors[:base] << "cannot delete submission that has already been paid"
      false
    end
  end

Answer 6

在Rails 5中你也可以：

def destroy
  quoted? && has_payments? ? self : super
end

submission.destroy # => submission
submission.destroyed? # => true/false

before_destroy回调不会停止记录被删除

6 个答案: