Question

我只是想让功能在特定时间运行，但我没有太多运气。让我来看看我所做的基本知识

#Gemfile

gem 'delayed_job_active_record'

#config/application.rb

module SampleApp
  class Application < Rails::Application
    # Settings in config/environments/* take precedence over those specified here.
    # Application configuration should go into files in config/initializers
    # -- all .rb files in that directory are automatically loaded.
    config.active_job.queue_adapter = :delayed_job
  end
end

#Ran from command line

[06.07.2017/15:38:02] user@ubuntu $ rails g delayed_job:active_record
Running via Spring preloader in process 72879
   identical  bin/delayed_job
       chmod  bin/delayed_job
      create  db/migrate/20170607203821_create_delayed_jobs.rb
[06.07.2017/15:38:21] user@ubuntu $ rake db:migrate
== 20170607203821 CreateDelayedJobs: migrating ================================
-- create_table(:delayed_jobs, {:force=>true})
   -> 0.0088s
-- add_index(:delayed_jobs, [:priority, :run_at], {:name=>"delayed_jobs_priority"})
   -> 0.0051s
== 20170607203821 CreateDelayedJobs: migrated (0.0142s) =======================

好的，所以我的视图上基本上有一个按钮调用post_now，然后调用我自定义模型中的函数。这是我所拥有的一个例子：

#posts_controller.rb
  def post_now
    set_post
    submit_post = SchedulePost.delay(run_at: 1.minute.from_now).schedule_post(@post)
    redirect_to posts_path, flash: {success: submit_post[1] }
  end

然后是SchedulePost类，它只是编写一个文本文件（暂时）

#models/schedule_post.rb
class SchedulePost < ActiveRecord::Base
    def self.schedule_post(post)
        command = "touch #{Rails.root}/public/test.txt"
        system(command)
        return
    end
end

所以我期待发生的事情是，当调用post_now时，它应该在1分钟后运行schedule_post函数，这应该只是写一个文本文件。如果我删除.delay(run_at: 1.minute.from_now)块并将schedule_post留在那里，那么它可以正常工作。所以我肯定在这里做了一些错误的延迟函数。

非常感谢任何帮助。

Answer 1

你需要启动delayed_job worker来执行延迟任务...在你的终端从app根目录执行

rake jobs:work

Delayed_job在RoR中不起作用或者我遗漏了什么？

1 个答案: