How to start/stop a function at specific time of the day?

Question

If I have a function that loops throughout the day denoted by:

def Show()

How can I run it so that at a particular time of the day it won't return anything?

In this instance say 13:50 to 14:20?

Gene Burinsky · Accepted Answer · 2021-11-17T15:26:52.090

3

If I understand the question correctly, you could just check the time using an if statement:

# import datetime module
from datetime import datetime as dt

# set to time bounds:
lower_bound = dt.time(13, 50)
upper_bound = dt.time(14, 20)

Then just check the time whenever the function runs:

now = dt.now().time()
if lower_bound < now < upper_bound:
     pass
if not (lower_boun < now < upper_bond):
     Show() #that returns something.

...or just put the if statements into the function itself

Answer improved by helpful comments of chepner and 0x5423

edited Nov 17 '21 at 15:26

answered Nov 16 '21 at 20:02

Gene Burinsky

1

Why use `dt.now` at all? `lower_bound = datetime.time(13, 50)` – chepner Nov 16 '21 at 20:06
1

Tip: Python lets you [chain](https://docs.python.org/3/reference/expressions.html#comparisons) comparison operators, so you can write that check as `if lower_bound < now < upper_bound:`. – 0x5453 Nov 16 '21 at 20:06
@chepner and 0x5453 - thanks for the suggestions. I incorporate them into the answer – Gene Burinsky Nov 16 '21 at 20:11
1

I would also just negate the condition, rather than use `pass` and an `else` clause. `if not (lower_bound < now < upper_bond): Show()`. Or `if now <= lower_bound or now >= upper_bound: Show()`. – chepner Nov 16 '21 at 20:13

score 3 · Answer 2 · answered Nov 16 '21 at 20:18

schedule is an excellent module for doing all of this. It even runs async without too much work.

schedule.every(1).hours.at("06:00").do(show).tag('show', 'guest')

and then stop the schedule at a specific time

schedule.every().day.at("10:30").clear('show')

2 Answers2