fetch start and end time for every job with delayed_job - ruby-on-rails

Is there any way i can fetch start and time for my every job. I am using delayed_job.

Depending on your setup, you could include actions to record the start time and end time of your job within the job itself.
class SomeJob < Struct.new(:param1, :param2)
def perform
start_time = Time.now
## Do Something
SomeModel.find(id).update_parameters({:start_time => start_time, :end_time => Time.now})
end
end
Might be easier than forking the repository and I am not crazy about the idea of keeping all of those jobs around, it would slow down the queue over time depending on load.

delayed_job has no ability to track the start time, duration or end time of a job. It also by default removes the table entry upon success.
You would have to fork the github version to and create a patch to track and record this information or utilise an external method ( http://helderribeiro.net/?p=87 uses monit ) to track this data (again which uses a forked version).

Related

ruby on rails background application to run jobs automaically at a time dynamically defined by users?

I have a use case where user schedules a 'command' from the web interface. The user also specifies the date and time the command needs to be triggred.
This is sequence of steps:
1.User schedules a command 'Restart Device' at May 31, 3pm.
2.This is saved in a database table called Command.
3.Now there needs to be a background job that needs to be triggered at this specified time to do something (make an api call, send email etc.)
4.Once job is executed, It is removed or marked done, until a new command is issued.
There could be multpile users concurrently performing the above sequence of steps.
Is delayed_job a good choice for above? I couldnt find an example as how to implement above using delayed job.
EDIT: the reason I was looking at delayed_job is because eventually I would need to leverage existing relational database
I would advise to use Sidekiq. With it you can use scheduled jobs to tell sidekiq when to perform the jobs.
Example :
MyWorker.perform_at(3.hours.from_now, 'mike', 1)
EDIT : worker example
#app/workers/restart_device_worker.rb
class RestartDeviceWorker
include Sidekiq::Worker
def perform(params)
# Do the job
# ...
# update in DB
end
end
see doc: https://blog.codeship.com/how-to-use-rails-active-job/
https://guides.rubyonrails.org/active_job_basics.html
If you are using Rails 5 then you have best option of ActiveJob(inbuilt feature)
Use ActiveJob
"Active Job – Make work happen later. Active Job is a framework for declaring jobs and making them run on a variety of queuing backends. These jobs can be everything from regularly scheduled clean-ups, to billing charges, to mailings. Anything that can be chopped up into small units of work and run in parallel, really."
Active Job has built-in adapters for multiple queuing backends (Sidekiq, Resque, Delayed Job and others). You just need to tell them.
Scenario: I want to delete my story after 24 hours(1 day). Then we do create a job named "StoriesCleanupJob". Call this job at the time of the creation of story like below
StoriesCleanupJob.set(wait: 1.day).perform_later(story)
It will call the Job after 1 day.
class StoriesCleanupJob < ApplicationJob
queue_as :default
def perform(story)
if story.destroy
#put your own conditions like update the status and all, whatever you want to perform.
end
end
end

How can calculate time from particular date/time instead current time in delayed job

I am using delayed job in rails. Now,
I am able to use run_at 1.minutes.from_now. and it is working fine. but as per my requirement time should be calculate from specific time like start_time in below code.
In controller level
Delayed::Job.enqueue DeviceJob.new(msg), :run_at => msg.start_time
In model level
class DeviceJob < Struct.new(:msg)
def perform
# something code
end
end
but in that case run_at will add in delayed job but on that time it will not process.
I don't have any other method in delayed job instead run_at.
I am using MongoDB(Nosql database)

Accessing rake task variables in controller and Scheduling rake tasks

I have a rake task send_emails which send e-mails to lot of people. I call this rake task from controller as mentioned in Rake in Background railscast. But I want to schedule this rake task to run at a particular date and time, which is not same for everyday (it's not a cron job). The date and time are set dynamically from a form.
For the above implemented rake task for sending emails, I want to show the status of the mailing process to the end-user. For instance, say there is a response object in the rake task which I can use as response.status,response.delivered?,response.address, etc. How can I access this object ( or any variable) in the rake file in my controller?
I don't want to use delayed_job but want to implement it's functionality of run_at and in_the_future. Also the whenever gem won't be able to solve my first problem coz I won't be able to pass date and time to it's scheduler.
First thing, calling rake task from controller is a bad practice. Ryan published that video at 2008 since that many better solution have came up. You shouldn't ignore it.
I suggest you to use delayed_job, it serves your needs in a great way. Since, if you want to invoke task dynamically, there should be some checker which will continuously check the desire field every second. Delayed job keep checking its database every time, you can use that.
Anyway,You can use something like this
def self.run_when
Scheduler.all.each do |s|
if d.dynamically_assigned_field < 1.second.ago
d.run_my_job!
d.status = "finished"
d.save
end
end
end
And, in model you can do something like this
def run_my_job!
self.status = "processing"
self.save
long_running_task
end
One thing also you should keep in mind that if too many workers/batch/cron job starts at run at same it will fight for resources and may enter into deadlock state. As per your server capacity, you should limit the running jobs.
Sidekiq is also a good option you can consider. Personally, i like sidekiq because it doesn't hit my database everytime , scales very effectively. It uses redis but it is expensive.
I would create new model for mail job, like this:
app/models/mail_job.rb
class MailJob
attr_accessible :email, :send_at, :delivered
scope :should_deliver, -> { where(delivered: false).where('send_at <= ?', Time.now) }
def should_deliver?
!delivered? && send_at <= Time.now
end
...
end
And use Sidekiq + Sidetiq, running every minute (or any other interval) and checking for mail jobs that should be delivered.
Hope this helps!

is there a way to run a job at a set time later, without cron, say a scheduled queue?

I have a rails application where I want to run a job in the background, but I need to run the job 2 hours from the original event.
The use case might be something like this:
User posts a product listing.
Background job is queued to syndicate listing to 3rd party api's, but even after original request, the response could take a while and the 3rd party's solution is to poll them every 2 hours to see if we can get a success acknowledgement.
So is there a way to queue a job, so that a worker daemon knows to ignore it or only listen to it at the scheduled time?
I don't want to use cron because it will load up a whole application stack and may be executed twice on overlapping long running jobs.
Can a priority queue be used for this? What solutions are there to implement this?
try delayed job - https://github.com/collectiveidea/delayed_job
something along these lines?
class ProductCheckSyndicateResponseJob < Struct.new(:product_id)
def perform
product = Product.find(product_id)
if product.still_needs_syndicate_response
# do it ...
# still no response, check again in two hours
Delayed::Job.enqueue(ProductCheckSyndicateResponseJob.new(product.id), :run_at => 2.hours.from_now)
else
# nothing to do ...
end
end
end
initialize job first time in controller or maybe before_create callback on model?
Delayed::Job.enqueue(ProductCheckSyndicateResponseJob.new(#product.id), :run_at => 2.hours.from_now)
Use the Rufus Scheduler gem. It runs as a background thread, so you don't have to load the entire application stack again. Add it to your Gemfile, and then your code is as simple as:
# in an initializer,
SCHEDULER = Rufus::Scheduler.start_new
# then wherever you want in your Rails app,
SCHEDULER.in('2h') do
# whatever code you want to run in 2 hours
end
The github page has tons of more examples.

Rails 3.1/rake - datespecific tasks without queues

I want to give my users the option to send them a daily summary of their account statistics at a specific (user given) time ....
Lets say following model:
class DailySummery << ActiveRecord::Base
# attributes:
# send_at
# => 10:00 (hour)
# last_sent_at
# => Time of the last sent summary
end
Is there now a best practice how to send this account summaries via email to the specific time?
At the moment I have a infinite rake task running which checks permanently if emails are available for sending and i would like to put the dailysummary-generation and sending into this rake task.
I had a thought that I could solve this with following pseudo-code:
while true
User.all.each do |u|
u.generate_and_deliver_dailysummery if u.last_sent_at < Time.now - 24.hours
end
sleep 60
end
But I'm not sure if this has some hidden caveats...
Notice: I don't want to use queues like resq or redis or something like that!
EDIT: Added sleep (have it already in my script)
EDIT: It's a time critical service (notification of trade rates) so it should be as fast as possible. Thats the background why I don't want to use a queue or job based system. And I use Monit to manage this rake task, which works really fine.
There's only really two main ways you can do delayed execution. You run the script when an user on your site hits a page, which is inefficient and not entirely accurate. Or use some sort of background process, whether it's a cron job or resque/delayed job/etc.
While your method of having an rake process run forever will work fine, it's inefficient because you're iterating over users 24/7 as soon as it finishes, something like:
while true
User.where("last_sent_at <= ? OR last_sent_at = ?", 24.hours.ago, nil).each do |u|
u.generate_and_deliver_dailysummery
end
sleep 3600
end
Which would run once an hour and only pull users that needed an email sent is a bit more efficient. The best practice would be to use a cronjob though that runs your rake task though.
Running a task periodically is what cron is for. The whenever gem (https://github.com/javan/whenever) makes it simple to configure cron definitions for your app.
As your app scales, you may find that the rake task takes too long to run and that the queue is useful on top of cron scheduling. You can use cron to control when deliveries are scheduled but have them actually executed by a worker pool.
I see two possibilities to do a task at a specific time.
Background process / Worker / ...
It's what you already have done. I refactored your example, because there was two bad things.
Check conditions directly from your database, it's more efficient than loading potential useless data
Load users by batch. Imagine your database contains millions of users... I'm pretty sure you would be happy, but not Rails... not at all. :)
Beside your code I see another problem. How are you going to manage this background job on your production server? If you don't want to use Resque or something else, you should consider manage it another way. There is Monit and God which are both a process monitor.
while true
# Check the condition from your database
users = User.where(['last_sent_at < ? OR created_at IS NULL', 24.hours.ago])
# Load by batch of 1000
users.find_each(:batch_size => 1000) do |u|
u.generate_and_deliver_dailysummery
end
sleep 60
end
Cron jobs / Scheduled task / ...
The second possibility is to schedule your task recursively, for instance each hour or half-hour. Correct me if I'm wrong, but do your users really need to schedule the delivery at 10:39am? I think that let them choose the hour is enough.
Applying this, I think a job fired each hour is better than an infinite task querying your database every single minute. Moreover it's really easy to do, because you don't need to set up anything.
There is a good gem to manage cron task with the ruby syntax. More infos here : Whenever
You can do that, you'll need to also check for the time you want to send at. So starting with your pseudo code and adding to it:
while true
User.all.each do |u|
if u.last_sent_at < Time.now - 24.hours && Time.now.hour >= u.send_at
u.generate_and_deliver_dailysummery
# the next 2 lines are only needed if "generate_and_deliver_dailysummery" doesn't sent last_sent_at already
u.last_sent_at = Time.now
u.save
end
end
sleep 900
end
I've also added the sleep so you don't needlessly hammer your database. You might also want to look into limiting that loop to just the set of users you need to send to. A query similar what Zachary suggests would be much more efficient than what you have.
If you don't want to use a queue - consider delayed job (sort of a poor mans queue) - it does run as a rake task similar to what you are doing
https://github.com/collectiveidea/delayed_job
http://railscasts.com/episodes/171-delayed-job
it stores all tasks in a jobs table, usually when you add a task it queues it to run as soon as possible, however you can override this to delay it until a specific time
you could convert your DailySummary class to DailySummaryJob and once complete it could re-queue a new instance of itself for the next days run
How did you update the last_sent_at attribute?
if you use
last_sent_at += 24.hours
and initialized with last_sent_at = Time.now.at_beginning_of_day + send_at
it will be all ok .
don't use last_sent_at = Time.now . it is because there may be some delay when the job is actually done , this will make the last_sent_at attribute more and more "delayed".

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