Can't seem to use the 'sched' module the way I want it to work

Question

The python project I'm working on, is a webpage that outputs covid data and news at a scheduled time. The project must use the sched module.

The issue I'm having it to do with schedule_covid_update() function. When calling the function. You have 2 required argument, the duration/time for the update to occur, the name of the update. And the 2 optional arguments, news and data. Which are used to tell schedule_covid_update() if both or just one update type is occurring at the scheduled time.

Now, if I were to call the function twice like this for example:

schedule_covid_update(4, "test update 2", news="news", data="data")
schedule_covid_update(0, "test update 1", news="news", data="data")

I would expect the data for "test update 1" to occur and return the data before "test update 2", regardless of the order I called schedule_covid_update(). But It doesn't, it schedules "test update 2".

Also when testing this code. You can just try some other function instead of updateData and updateNews. As I can't give the code for the functions.

scheduler.py:

import time, sched
from datetime import datetime
from covid_data_handler import *


s = sched.scheduler(time.time, time.sleep)
def schedule_covid_update (update_interval, update_name, news="", data=""):
    print("update name: " + update_name)

    if(news=="news" and data=="data"):
        #s.enter(timeDuration(update_interval),1, updateData)
        #s.enter(timeDuration(update_interval),1, updateNews)
        s.enter(update_interval,1, updateData)
        s.enter(update_interval,1, updateNews)
        print("Updating news and data")
    elif(news=="news" and data!="data"):
        #s.enter(timeDuration(update_interval),1, updateNews)
        s.enter(update_interval,1, updateNews)
        print("Updating news only")
    elif(news!="news" and data=="data"):
        #s.enter(timeDuration(update_interval),1, updateData)
        s.enter(update_interval,1, updateData)
        print("Updating data only")
        #scheduler.run(blocking=False)
    else:
        print("No Update")

    s.run()


schedule_covid_update(4, "test update 2", news="news", data="data")
schedule_covid_update(0, "test update 1", news="news", data="data")

Please see: [Why is Can someone help me? not an actual question?](https://meta.stackoverflow.com/questions/284236/why-is-can-someone-help-me-not-an-actual-question) for more details. — itprorh66, Jul 19 '22 at 15:13
Also see [Under what circumstances may I add "urgent" or other similar phrases to my question, in order to obtain faster answers?](https://meta.stackoverflow.com/questions/326569/under-what-circumstances-may-i-add-urgent-or-other-similar-phrases-to-my-quest) — martineau, Jul 19 '22 at 20:02

score 1 · Answer 1 · answered Jul 19 '22 at 15:13

1

Take a look at the docs for scheduler.run . With s.run() inside of your function, it blocks forever and never reaches the second scheduled task. Schedule first, then run the scheduler independently, outside of the scheduling functions.

answered Jul 19 '22 at 15:13

Dominik Stańczak

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Can't seem to use the 'sched' module the way I want it to work

1 Answers1