Question

尝试使用delayed_job对作业进行排队，如下所示：

Delayed::Job.enqueue(BackgroundProcess.new(current_user, object))

当我打印出来时，

current_user和object不是nil。奇怪的是，有时刷新页面或再次运行命令有效！

以下是异常跟踪：

  Delayed::Backend::ActiveRecord::Job Columns (44.8ms)   SHOW FIELDS FROM `delayed_jobs`

TypeError (wrong argument type nil (expected Data)):
  /Users/.rvm/rubies/ruby-1.9.1-p378/lib/ruby/1.9.1/yaml.rb:391:in `emit'
  /Users/.rvm/rubies/ruby-1.9.1-p378/lib/ruby/1.9.1/yaml.rb:391:in `quick_emit'
  /Users/.rvm/rubies/ruby-1.9.1-p378/lib/ruby/1.9.1/yaml/rubytypes.rb:86:in `to_yaml'
  vendor/plugins/delayed_job/lib/delayed/backend/base.rb:65:in `payload_object='
  activerecord (2.3.9) lib/active_record/base.rb:2918:in `block in assign_attributes'
  activerecord (2.3.9) lib/active_record/base.rb:2914:in `each'
  activerecord (2.3.9) lib/active_record/base.rb:2914:in `assign_attributes'
  activerecord (2.3.9) lib/active_record/base.rb:2787:in `attributes='
  activerecord (2.3.9) lib/active_record/base.rb:2477:in `initialize'
  activerecord (2.3.9) lib/active_record/base.rb:725:in `new'
  activerecord (2.3.9) lib/active_record/base.rb:725:in `create'
  vendor/plugins/delayed_job/lib/delayed/backend/base.rb:21:in `enqueue'

Answer 1

我猜这是因为您将对象作为参数发送到您的作业（至少我认为current_user和对象实际上是对象而不是id）。改为发送id，然后在开始执行时加载对象。

例如：

Delayed::Job.enqueue(BackgroundProcess.new(current_user.id, object.id))

class BackgroundProcess < Struct.new(:user_id, :object_id)
  def perform
    @current_user = User.find(user_id)
    @object = Object.find(object_id)

    ...
  end
end

这样，将ActiveRecord序列化到数据库中不会有任何问题，并且您将始终在作业运行时加载最新的更改。

Answer 2

也遇到同样的问题。我仍然不知道是什么导致它，但由于某种原因克隆对象似乎解决了它

u = User.find 123
u.to_yaml
=> TypeError: wrong argument type nil (expected Data)
u.clone.to_yaml
=> works like normal

非常令人沮丧。最好知道根本原因，但如果你绝望，这可能会有所帮助。

delayed_job奇怪的例外

2 个答案: