Rails 4.2:使用deliver_later和无表格模型

Question

我正在尝试使用Rails 4.2的deliver_later方法设置联系表单.但是,我只能让deliver_now工作,因为deliver_later正在尝试序列化我的对象并且每次都失败.

这是我的设置:

messages_controller.rb

class MessagesController < ApplicationController
  def new
    @message = Message.new
  end

  def create
    @message = Message.new(params[:message])
    if @message.valid?
      ContactMailer.contact_form(@message).deliver_later
      redirect_to root_path, notice: "Message sent! Thank you for contacting us."
    else
      render :new
    end
  end
end

contact_mailer.rb

class ContactMailer < ApplicationMailer
  default :to => Rails.application.secrets['email']

  def contact_form(msg)
    @message = msg
    mail(:subject => msg.subject, from: msg.email)
  end
end

message.rb

class Message
    include ActiveModel::Model
    include ActiveModel::Conversion

    ## Not sure if this is needed ##
    include ActiveModel::Serialization

    extend ActiveModel::Naming

    attr_accessor :name, :subject, :email, :body

    validates_presence_of :email, :body
    validates_format_of :email, with: /\A([^\s]+)((?:[-a-z0-9]\.)[a-z]{2,})\z/i
    validates_length_of :body, :maximum => 1000

    def initialize(attributes = {})
      attributes.each { |name, value| send("#{name}=", value) }
    end

    ## Not sure if this is needed ##
    def attribtues
      {'name' => nil, 'subject' => nil, 'email' => nil, 'body' => nil}
    end
end

我打电话时遇到的错误ContactMailer.contact_form(@message).deliver_later是:

ActiveJob::SerializationError in MessagesController#create 

Unsupported argument type: Message
Extracted source (around line #10): 
if @message.valid?
  ContactMailer.contact_form(@message).deliver_later
  redirect_to root_path, notice: "Message sent! Thank you for contacting us."
else
  render :new

理想情况下,我希望这是一个后台进程.我将尽快添加类似Sidekiq的东西,但我认为最好事先修复此序列化问题.

任何帮助表示赞赏!谢谢 :)

Answer 1

为了使用您的类ActiveJob(deliver_later委托给它),它需要能够通过其ID唯一地标识对象.此外,它需要在反序列化时通过ID找到它(在邮件程序/作业中不需要手动反序列化).

class Message
  ...
  include GlobalID::Identification
  ...

  def id
    ...
  end

  def self.find(id)
    ...
  end
end

ActiveRecord会为你提供这些方法,但由于你没有使用它,你需要自己实现它.由你来决定你想要存储记录的位置,但说实话,我认为通过使用ActiveRecord和下面的表格会更好.

Answer 2

一个简单的解决方案,避免必须使用ActiveRecord支持对象或创建不必要的表:

您也可以将消息参数传递给contact_form方法,然后在该方法中初始化Message对象,而不是将Message对象传递给contact_form方法.

这将解决问题而无需创建表,因为您正在初始化延迟作业工作者的内存空间中的对象.

例如:

messages_controller.rb

MessagesController < ApplicationController
    def new
        @message = Message.new
    end

    def create
        @message = Message.new(params[:message])

        if @message.valid?
            ContactMailer.contact_form(params[:message]).deliver_later
            redirect_to root_path, notice: "Message sent! Thank you for contacting us."
        else
            render :new
        end
    end
end

contact_mailer.rb

class ContactMailer < ApplicationMailer
    default :to => Rails.application.secrets['email']

    def contact_form(msg_params)
        @message = Message.new(msg_params)
        mail(:subject => msg.subject, from: msg.email)
    end
end