How do I ensure that a the same job can't run twice? (unique job / avoid duplicates)

[rgaufman]

I am looking for something like this for good_job: https://github.com/mhenrixon/sidekiq-unique-jobs

To be able to do specifically the equivalent of this:

sidekiq_options queue: :default, lock: :until_executed

Is this the equivalent with good_job? <

 good_job_control_concurrency_with(total_limit: 1, key: arguments)

Or am I misunderstanding?

[bensheldon]

Exactly. GoodJob's ActiveJob Concurrency extension can provide equivalent functionality to sidekiq-unique-jobs

This would be the equivalent of lock: :until_executed

class MyJob < ApplicationJob
  include GoodJob::ActiveJobExtensions::Concurrency

  good_job_control_concurrency_with(total_limit: 1)
  
  # ....
end

[rgaufman]

Just to confirm:

1. This allow many of the same job to run concurrently with different arguments
2. But it won't allow queuing a new job if one is already running with the same arguments

Is that right?

[bensheldon]

Concurrency is limited based on whatever the key: value is. By default the key is the name of the class that the concurrency control is declared within, but you can use the lambda to make it based on anything (queue name, job arguments, some database call, anything) by using the Proc that returns a String that is compared against.

[rgaufman]

Ah. ok, so then this should work and allow multiples of the same job, as long as arguments are different? (whatever the arguments are):

good_job_control_concurrency_with(total_limit: 1, key: arguments)

[bensheldon]

That exact code won't execute because arguments only exist in the context of a particular instance of a job. You'd need to stringify the arguments within a proc. e.g.

# an example scalpel based on your specific method signature
good_job_control_concurrency_with(total_limit: 1, key: -> { arguments[0] })

# or with a hammer
good_job_control_concurrency_with(total_limit: 1, key: -> { arguments.to_json })

[rgaufman]

Why is .to_json necessary? - can I specify just arguments in the lambda?

[bensheldon]

You'll get whatever arguments.to_s stringifies to. Which is probably a consistent marshalling of the object, but I dunno. Give it a shot.

[rgaufman]

Ah, understood, thank you!