在作业完成后更新控制器

Question

我有一个控制器动作，它触发一个延迟的作业，它有一个after_perform钩子，在这个钩子中触发了另一个延迟的作业。

我需要在所有作业完成后渲染一个视图，并向用户显示一个微调器，同时作业仍在运行。

这些作业如下所示：

class DuplicateShipmentJob < ApplicationJob
  after_perform do |job|
    shipment = job.arguments.first[:shipment]
    shipment_ids = Shipment.where(public_id: shipment.public_id).map(&:id)
    bulk_shipment_process = BulkShipmentProcess.new
    bulk_shipment_process.shipment_ids = shipment_ids
    bulk_shipment_process.user = shipment.user

    if bulk_shipment_process.save
      Shipment.where(public_id: shipment.public_id).each do |shipment|
        shipment.bulk_shipment_process_id = bulk_shipment_process.id
      end
      BulkShipmentsProcessor.delay.process_bulk_shipments(args)
    end
   end

 def perform(shipment:, label_count:)
   label_count.times do
     duplicated_shipment = shipment.dup
     duplicated_shipment.package = shipment.package.dup
     duplicated_shipment.save!
   end
  end
 end

做这件事最好的方法是什么？

我在向数据库发出ajax请求和设置动作电缆来监听作业结果之间犹豫不决。

控制器看起来像这样：

class ShipmentsController < ApplicationController
 def replicate_shipment
  @shipment = build_shipment
  @shipment.save!
  DuplicateShipmentJob.perform_later(
    shipment: @shipment,
    label_count: @shipment.create_multiple_shipments.to_i - 1,
  )
  # here i need to wait until the process instance appears in the DB
  @bulk_shipment_process = BulkShipmentProcess.where(id: Shipment.last.bulk_shipment_process_id) 
  redirect_to bulk_shipment_process_path(@bulk_shipment_process)
 end
end

Answer 1

我首先建议将作业存储在数据库中，并循环遍历与触发它的用户相关的作业，并识别它。

提交作业后，触发旋转器，然后尝试硬编码一个从数据库检索相关作业的开放循环，并检查其状态是否为done

        loop do
          status = DelayedJob.find(JOB_ID)
          case status
          when 'DONE' then return true
          when 'PROCESSING' then sleep(10)
          else raise(JobRunFailed, "Job run failed with: #{status}")
          end
        end

，取下微调控件并输出您想要的相关消息。

当然，如果您愿意，也可以将开环逻辑硬编码在应用程序的前端。

因此，简单回顾一下这些步骤：

1. 标识数据库中存储的相关作业，并在此处触发
2. 旋转器，您可以在此处通过标识的作业重定向到bulk_shipment_process_path
3. Loop，直到循环中断后将其标记为

或

1. 相应的接口。

第3步和第4步实际上可能只是与前端相关的步骤，与ShipmentsController无关。

Answer 2

我最终得到了以下解决方案。

我将BulkShipmentProcess的实例化移到了控制器，并在保存它之后，重定向到此实例的show操作，并触发了第一个作业，将该实例作为作业参数传递。

在作业内部，触发第二个作业，并将一些属性分配给BulkShipmentProcess实例。

当在bulk_shipment_process/id页面上时，JS正在为实例轮询DB，以获得分配的新属性，并在成功时呈现这些属性。

控制器：

class ShipmentsController < ApplicationController
  def replicate_shipment
    @shipment = build_shipment
    @shipment.save!
    @bulk_shipment_process = BulkShipmentProcess.new()
    @bulk_shipment_process.save
    if @bulk_shipment_process.id?
       DuplicateShipmentJob.perform_later(@bulk_shipment_process)
       redirect_to bulk_shipment_process_path(@bulk_shipment_process)
    else
      flash.now[:error] = "error"
      render :new
    end
  end
end

作业：

class DuplicateShipmentJob < ApplicationJob
 def perform(bulk_shipment_process:)
  # do something
  # assign new attributes to the instance of interest
  # trigger second delayed job upon instance of interest
 end
end

轮询功能：

currentStatus = (element) ->
  element.find("div").data("status")

statusFinished = (status) ->
  status.startsWith("finish") ||
    status.startsWith("fail") ||
    status.startsWith("unknown") ||
    status.startsWith("stuck")

statusUpdate = (element, done_action) ->
  $.ajax
    url: element.data("target")
    success: (data) ->
      element.html data
      status = currentStatus element

      if statusFinished(status)
        console.log "done_action " + done_action
        if done_action == "reload"
          console.log "reload"
          location.reload()
      else
        setTimeout (->
          statusUpdate(element, done_action)
          return
        ), 5000
      return
  return

$ ->
  $('[data-controller="status-update"]').each (index) ->
    target = $(this).data("target")
    status = currentStatus $(this)
    done_action = $(this).data("done")

    unless statusFinished(status)
      statusUpdate($(this), done_action)
    return