Rails 后台工作者总是第一次失败，第二次工作

本文介绍了Rails 后台工作者总是第一次失败，第二次工作的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我有一个在保存对象后触发的 sidekiq 工作器:

I have a sidekiq worker that fires after an object is saved:

class Post < ActiveRecord::Base attr_accessible :description, :name, :key after_save :process def process ProcessWorker.perform_async(id, key) if key.present? end def secure_url key.match(/(.*\/)+(.*$)/)[1] end def nonsecure_url key.gsub('https', 'http') end end

worker 如下所示(它还没有完成……只是在测试):

The worker looks like the following (it's not complete yet... just testing):

class ProcessWorker include Sidekiq::Worker def perform(id, key) post = Post.find(id) puts post.nonsecure_url end end

奇怪的是，每次 worker 触发时，它最初都会失败并显示以下错误:

Oddly enough, every time the worker fires, it initially fails with the following error:

undefined method `gsub' for nil:NilClass

但是，当工人稍后重试时 - 它总是成功.

But then, when the worker retries a bit later – it always succeeds.

感觉好像有些东西没有在它应该初始化的时候初始化......但我似乎无法追踪它.

It really feels as though something isn't initializing when it should... but I can't seem to track it down.

推荐答案

确保 process always true - 这是需要的一部分回调链.

Ensure that process always true - which is needed as part of the callback chain.

所以改写成这样:

def process ProcessWorker.perform_async(id, key) if key.present? true end

当执行 ActiveRecord 回调链时，任何返回布尔值的方法都会影响该链 - false 将取消回调链.

When the ActiveRecord callback chain is being executed any methods that return a boolean will affect the chain - a false will cancel the callback chain.