我有一个函数被标记为由延迟作业异步处理:
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
当我运行这个函数时,我会传入一个代码块:
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
在没有使用 delayed_job 运行时,这个代码能正常工作,但当使用它时,我会得到以下错误
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 ...
看起来没有获取到传入的代码块。这样做不正确吗?还是我应该寻找其他方法?