Schedule a job or function like enqueue but for once and at specific time or day

Frappe Framework
1

I am looking schedule some function to be execute at a specific time,
@Duradundi_hadimani @DaizyModi @rmehta

Looking for a help
Thanks In advance.

2

Try Scheduler events in hooks.py
https://docs.frappe.io/framework/user/en/api/background_jobs

You can put in a custom time of day using CRON format as well.

3

One option is to set scheduler for specific time. And from python create empty flag file .alreadyran that will skip run after that file flag is created. Other, cleaner option, is to use ‘at’ command from linux.

4

Hi @nikkothari22 , Thanks for Reply

frappe.enqueue

I tried using this but cant specify the day or time of execution and hooks is not possible in my case as I want it to create dynamically.

5

Hi @MrKovalski , Thanks for Reply

setting scheduler is not a proper solution for me , can you please provide some reference for at command from Linux and how can I use it.

6

Check rq-scheduler. You will need to start it as a process. One more service for rqscheduler will have to run along with other bench services.

In brief it will be:

  • Add python lib to your custom app’s dependencies (pyproject.toml)
  • Start the process using ./env/bin/rqscheduler --url=$(jq -r ".redis_queue" sites/common_site_config.json)
  • Use as per documentation from your custom app code, e.g. scheduler.enqueue_in(timedelta(minutes=10), count_retweets, tweet_id)
  • you may have to write your own command: bench scheduler just like existing command bench worker, refer frappe/commands/scheduler.py and start_worker from frappe/utils/background_jobs.py