Question

我正在编写一个Rails 4.2.0应用程序，我需要发送电子邮件。我被告知Que gem用于管理后台工作。我已经做了所有事情，因为安装和使用列出了here。

此外，我在application.rb这些行中指定了：

    # For https://github.com/chanks/que
    config.active_record.schema_format = :sql
    config.active_job.queue_adapter = :que

我的工作看起来像send_welcome_message.rb：

class SendWelcomeEmail < Que::Job
  # Default settings for this job. These are optional - without them, jobs
  # will default to priority 100 and run immediately.
  @priority = 10
  @run_at = proc { 1.minute.from_now }

  def run(user_id, options)

    @user = User.find(user_id)
    UserMailer.welcome_email(@user).deliver_now

    # Destroy the job.
    destroy
  end
end

运行rails s命令后，我的控制台将填充以下消息：

{
   "lib":"que",
   "hostname":"...",
   "pid":13938,
   "thread":69925811873800,
   "event":"job_unavailable"
}

当我在控制器中将这样的工作排入队列时

SendWelcomeEmail.enqueue 20, priority: 100

并刷新页面，我总是得到以下错误（尽管我可以发送消息是同步方式而不使用que）：

    {
   "lib":"que",
   "hostname":"...",
   "pid":13938,
   "thread":69925811873800,
   "event":"job_errored",
   "error":{
      "class":"ArgumentError",
      "message":"wrong number of arguments (1 for 2)"
   },
   "job":{
      "queue":"",
      "priority":100,
      "run_at":"2015-06-22T01:59:45.187+03:00",
      "job_id":11,
      "job_class":"SendWelcomeEmail",
      "args":[
         20
      ],
      "error_count":2
   }
}

当我在第二个终端中打开rails console并输入Que.worker_states（it's written here and should return information about every worker in the system）时，我得到[]。

我认为我没有产生任何工人。我对吗？以及如何解决它？

更新

在que log中发现错误：

wrong number of arguments (1 for 2)
/home/username/train/project/app/jobs/send_welcome_email.rb:8:in `run'
/home/username/.rbenv/versions/2.1.2/lib/ruby/gems/2.1.0/gems/que-0.10.0/lib/que/job.rb:15:in `_run'
/home/username/.rbenv/versions/2.1.2/lib/ruby/gems/2.1.0/gems/que-0.10.0/lib/que/job.rb:99:in `block in work'
/home/username/.rbenv/versions/2.1.2/lib/ruby/gems/2.1.0/gems/que-0.10.0/lib/que/adapters/active_record.rb:5:in `block in checkout'
/home/username/.rbenv/versions/2.1.2/lib/ruby/gems/2.1.0/gems/activerecord-4.2.0/lib/active_record/connection_adapters/abstract/connection_pool.rb:292:in `with_connection'
/home/username/.rbenv/versions/2.1.2/lib/ruby/gems/2.1.0/gems/que-0.10.0/lib/que/adapters/active_record.rb:34:in `checkout_activerecord_adapter'
/home/username/.rbenv/versions/2.1.2/lib/ruby/gems/2.1.0/gems/que-0.10.0/lib/que/adapters/active_record.rb:5:in `checkout'
/home/username/.rbenv/versions/2.1.2/lib/ruby/gems/2.1.0/gems/que-0.10.0/lib/que/job.rb:82:in `work'
/home/username/.rbenv/versions/2.1.2/lib/ruby/gems/2.1.0/gems/que-0.10.0/lib/que/worker.rb:78:in `block in work_loop'
/home/username/.rbenv/versions/2.1.2/lib/ruby/gems/2.1.0/gems/que-0.10.0/lib/que/worker.rb:73:in `loop'
/home/username/.rbenv/versions/2.1.2/lib/ruby/gems/2.1.0/gems/que-0.10.0/lib/que/worker.rb:73:in `work_loop'
/home/username/.rbenv/versions/2.1.2/lib/ruby/gems/2.1.0/gems/que-0.10.0/lib/que/worker.rb:17:in `block in initialize'

第8行是：

def run(user_id, options)

解

现在它的'工作。我已从application.rb删除了适配器配置，而不是

SendWelcomeEmail.enqueue 20, priority: 100

写

@user = User.find(20)
SendWelcomeEmail.enqueue  @user.id, :priority => 100

现在它的确有效。在第二个变体中有趣的事情是将相同的值传递给函数。仍有错误消息表示run只获得了1个参数 - 20。

Answer 1

阅读que gem，看起来enqueue方法将priority关键字视为一种特殊情况：https://github.com/chanks/que/blob/master/lib/que/job.rb#L31

因此，您的run方法仅传递第一个参数。 priority关键字被que吞没。

将run方法更改为

  def run(user_id)

应该解决您的问题。

使用后台作业管理器gem执行作业的错误称为Que

1 个答案: