Passing a Block to a delayed_job

Question

I have a function that is marked to be handled asynchronously by delayed_job:

class CapJobs
  def execute(params, id)
    begin
      unless Rails.env == "test"
        Capistrano::CLI.parse(params).execute!
      end
    rescue
      site = Site.find(id)
      site.records.create!(:date => DateTime.now, :action => "Task Failure: #{params[0]}", :type => :failure)    
      site.save
    ensure
      yield id
    end 
  end   
  handle_asynchronously :execute
end 

When I run this function I pass in a block:

 capjobs = CapJobs.new
 capjobs.execute(parameters, @site.id) do |id|
   asite = Site.find(id)
   asite.records.create!(:date => DateTime.now, :action => "Created", :type => :init)
   asite.status = "On Demo"
   asite.dev = true
   asite.save
 end

This works fine when run without delayed_job but when run with it I get the following error

2012-08-13T09:24:36-0300: [Worker(delayed_job host:eagle pid:12089)] SitesHelper::CapJobs#execute_without_delay failed with LocalJumpError: no block given (yield) - 0 failed attempts
2012-08-13T09:24:36-0300: [Worker(delayed_job host:eagle pid:12089)] PERMANENTLY removing SitesHelper::CapJobs#execute_without_delay because of 1 consecutive failures.
2012-08-13T09:24:36-0300: [Worker(delayed_job host:eagle pid:12089)] 1 jobs processed at 0.0572 j/s, 1 failed ...

It seems not to pick up the block that was passed in. Is this not the correct way of doing this or should I find a different method?

Answer 1

I found a method of doing this. It is kind of hacky but works well. I found this article that talks about creating a SerializableProc class. If I pass this to the function then everything works great.