在特定时间启动和停止 Python 脚本

Question

I am writing a script which will record the user activity in a certain period of time every day. The periods are taken from a JSON response of an API, like this:

[{"working_day": "mon", "work_start_at": "09:00", "work_end_at": "18:00"},
{"working_day": "tue", "work_start_at": "09:00", "work_end_at": "15:00"}]

Let's assume that I can parse these strings in a datetime() format. I want to run my function accordingly to these periods and stop my function after "work_end_at". I found numerous example of how to run after certain amount of seconds, how to run every day (Python lib schedule) and examples with bash crontab. But nothing of this works for me, because all I want is to start the script at a specific time and stop at a specific time. And then, run again when the next "shift" comes.

def start():
    time_periods = working_periods_table.find_all()
    today = datetime.datetime.now().isoweekday()
    for day in time_periods:
        if day["day"] == today:
            start_time = day["work_start_at"]
            end_time = day["work_end_at"]

    schedule.every().week.at(start_time).do(record_activity(idle_time=300))

Answer 1

If I've understood what you're asking, you could use while loops. Have it periodically check the current time with the times for the shift beginning and end. Below is a guide of what I mean, but I'm not sure the comparisons will work.

from time import sleep

def start():
    time_periods = working_periods_table.find_all()
    today = datetime.datetime.now().isoweekday()
    for day in time_periods:
        if day["day"] == today:
            start_time = day["work_start_at"]
            end_time = day["work_end_at"]

    while True:
        if datetime.datetime.now() >= start_time:
            while True:
                schedule.every().week.at(start_time).do(record_activity(idle_time=300))
                if datetime.datetime.now() >= end_time:
                    break
            break
        else:
            sleep(100)

Answer 2

I have some way to start my process at certain time but can not stop at certain time, I think you might means after running at 1 hour, even it is not finished yet, still stop the program. in this case you might can use while loop

take a look with my scripts first this might helpful

import time
from time import strftime, localtime
import datetime

now = datetime.datetime.now()
year = int(now.strftime("%Y"))
month = int(now.strftime("%m"))
day = int(now.strftime("%d"))
hour = int(now.strftime("%H"))

start_time_t = datetime.datetime(year=year, month=month, day=day, hour=22, minute=0, second=0)
waiting_time = time.mktime(start_time_t.timetuple()) - time.time()
if waiting_time >= 0:
    print('the program will be start after ' + time.strftime("%H hours and %M minutes", time.gmtime(waiting_time)))
    time.sleep(waiting_time)
else:
    start_time_t = datetime.datetime(year=year, month=month, day=day + 1, hour=00, minute=0, second=0)
    waiting_time = time.mktime(start_time_t.timetuple()) - time.time()
    print('the program will be start after ' + time.strftime("%H hours and %M minutes", time.gmtime(waiting_time)))
    time.sleep(waiting_time)

It will runs my program every day at 10:00 PM, if pass 10:00 PM the program will runs on the beginning of text day.

Answer 3

If you want to run your script at specific time intervals , I will suggest the code below which I have used on my own script. So simple to implement.

import datetime
now = datetime.datetime.now()

print(now.year, now.month, now.day, now.hour, now.minute, now.second)
# 2015 5 6 8 53 40


while True:
    now = datetime.datetime.now()
    if 14 <= now.hour and 23=> now.hour:
        # put your code here 

Note that in this piece of code I consider 2 PM until 11 PM. And more important you have to get the current hour persistently in a loop.

If you have a specific day in your mind get your current day out of the loop and compare it to your get value !! It works for me.