If return error, try again in 1 minute, if another return error occurs, try again in 2 minutes, and-CodePudding

To try again in 1 minute, I use sleep(60):

import requests
from time import sleep

headers = {
    "User-Agent": "Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/97.0.4692.99 Safari/537.36"
    }

def part_1():
    url = f'https://errorerrorerror.com'
    response = requests.get(url, headers=headers, timeout=1).json()
    return response

def main():
    print('Start Run')
    try:
        part_1()
    except:
        sleep(60)
        main()

if __name__ == '__main__':
    main()

So that I can try again when there is an error, I add in except the call again to main().

But I would like to add 1 more minute to each attempt, for example:

There was an error: try again in 1 minute
There was another error: try again in 2 minutes
There was another error: try again in 3 minutes

And so on. How should I proceed to make this possible?

CodePudding user response：

Avoid recursion in a case like this where there is no upper bound to the amount of times the function will recurse, as you will eventually get a Stack Overflow Error

def main():
    sleep_time = 60
    while True:
        print('Start Run')
        try:
            part_1()
            break
        except:
            sleep(sleep_time)
            sleep_time  = 60

if __name__ == '__main__':
    main()

CodePudding user response：

pause = 60
while True:
    try:
        part_1()
    except:
        sleep(pause)
        pause  = 60
    else:
        break