Question

我使用带有ActiveJob和sidekiq的rails作为后端。当用户进入页面sidekiq创建长期后台任务时，如何在任务完成时注意到用户（通过在网页上呈现部分）？

Rails和sidekiq作为不同的进程工作。这个事实让我很困惑，我不明白如何使用后台工作处理完成状态。

Answer 1

ActiveJob提供after_perform回调，根据文档的工作方式如下：

class VideoProcessJob < ActiveJob::Base
  queue_as :default

  after_perform do |job|
    UserMailer.notify_video_processed(job.arguments.first)
  end

  def perform(video_id)
    Video.find(video_id).process
  end
end

因此，您无需担心直接与Sidekiq或任何其他排队后端集成，请与ActiveJob对话：）

Answer 2

我在这种情况下的做法是：

添加sidekiq-status，以便按ID跟踪后台作业。

在创建后台作业的客户端调用中，返回新创建的作业ID。

class MyController < ApplicationController

  def create
    # sidekiq-status lets us retrieve a unique job ID when
    # creating a job
    job_id = Workers::MyJob.perform_async(...)

    # tell the client where to find the progress of this job
    return :json => {
      :next => "/my/progress?job_id={job_id}"
    }
  end

end

使用该作业ID轮询服务器上的“进度”端点。此端点获取作业的作业进度信息并将其返回给客户端。

class MyController < ApplicationController

  def progress
    # fetch job status from sidekiq-status
    status = Sidekiq::Status::get_all(params[:job_id])

    # in practice, status can be nil if the info has expired from
    # Redis; I'm ignoring that for the purpose of this example

    if status["complete"]
      # job is complete; notify the client in some way
      # perhaps by sending it a rendered partial
      payload = {
        :html => render_to_string({
          :partial => "my/job_finished",
          :layout => nil
        })
      }
    else
      # tell client to check back again later
      payload = {:next => "/my/progress?job_id={params[:job_id]}"}
    end

    render :json => payload
  end

end

如果客户端看到作业已完成，则可以显示消息或采取下一步所需的步骤。

var getProgress = function(progress_url, poll_interval) {
  $.get(progress_url).done(function(progress) {
    if(progress.html) {
      // job is complete; show HTML returned by server
      $('#my-container').html(progress.html);
    } else {
      // job is not yet complete, try again later at the URL
      // provided by the server
      setTimeout(function() {
        getProgress(progress.next, poll_interval);
      }, poll_interval);
    }
  });
};
$("#my-button").on('click', function(e) {
  $.post("/my").done(function(data) {
    getProgress(data.next, 5000);
  });
  e.preventDefault();
});

警告：该代码是说明性的，并且缺少您应该处理的事项，例如错误处理，防止重复提交等等。

后台任务完成后如何通知用户？

2 个答案: