使用后台作业管理器gem执行作业的错误称为Que

Question

我正在编写一个Rails 4.2.0应用程序，我需要提供电子邮件。我曾被建议Que gem来管理后台工作。我已经完成了所有的安装和使用列表here。

而且我在application.rb已经指定了这些行：

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

我的工作看起来像这样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命令我的控制台中填充这些消息后：

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

当我像我的工作排队像这在控制器

SendWelcomeEmail.enqueue 20, priority: 100

，并刷新页面，我收到以下错误所有的时间（尽管我可以发送消息是同步的方式，而不使用阙）：

{ 
    "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 
    } 
} 

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

我认为我没有工人产卵。我对吗？以及如何解决它？

UPDATE

在阙日志

实测值误差：8

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' 

线是：

def run(user_id, options) 

SOLUTION

现在我ts'工作。我从application.rb删除适配器配置和在

SendWelcomeEmail.enqueue 20, priority: 100

地方写

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

现在，它的作品。在第二种变体中，有趣的是相同的值被传递给函数。仍然错误消息说，run只获得1个参数 - 20。

Answer 1

读que宝石，它看起来像enqueue方法治疗priority关键字作为一个特殊情况：https://github.com/chanks/que/blob/master/lib/que/job.rb#L31

所以，你run方法仅通过了第一个参数。 priority关键字被que吞噬。

改变你run方法

def run(user_id) 

应该解决您的问题。