On 15 Aug., 14:19, Frederick Cheung <[email protected]> wrote: > On Aug 15, 1:10 pm, Genji <[email protected]> wrote: > Well the problem WAS that it didn't do anything, the worker just told me:
[Worker(host:Sascha-Gundels-MacBook-Pro.local pid:961)] 2 jobs processed at 96.8147 j/s, 2 failed ... But it changed now. All I did was setting up /config/initializers/delayed_job_config.rb to: Delayed::Worker.destroy_failed_jobs = false Delayed::Worker.sleep_delay = 60 Delayed::Worker.max_attempts = 3 Delayed::Worker.max_run_time = 5.minutes and restart the worker ^^' Sorry for asking - it works now.. > > > > > > > > > Hi all, > > > I wrote a simple code to set up the "pit_id" from a village table from > > 1 to 2, from 2 to 3 and so on. > > It works fine without .delay . But i need it so whats wrong with that > > code? > > > #village_controller > > def post_info > > @village = Village.find(params[:id]) > > @village.delay.upgrade > > flash[:notice] = "Upgrading" > > redirect_to village_url(@village) > > end > > > #village.rb > > def upgrade > > sleep 3 > > update_attribute(:pit_id, pit_id + 1) > > end > > > I've heard that there could be problems with the variables when using > > delay ? > > In what way does it not work ? > > Fred > > > > > > > > > I already tried @village.delay.upgrade(@village) -- You received this message because you are subscribed to the Google Groups "Ruby on Rails: Talk" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/rubyonrails-talk?hl=en.
