Python Requests try/except continue/break loop

Question

For various reasons I need to put a Requests call inside a try/except/retry loop, rather mounting a retry condition to a requests session. Expected behaviour is that if a request has been successful, the loop breaks and the code stops. Actual behaviour though is that it repeats the loop from start to finish, with the break statement seemingly having no effect:

import traceback import requests import time for i in range(0, 15): while True: try: headers ={ 'authority': 'www.wikipedia.org', 'method': 'GET', 'path': '/wikipedia.org', 'scheme': 'https', 'accept': '*/*', 'accept-language': 'en-GB,en-US;q=0.9,en;q=0.8', 'referer': 'https://google.com', 'sec-fetch-dest': 'empty', 'sec-fetch-mode': 'cors', 'sec-fetch-site': 'same-origin', 'user-agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/87.0.4280.66 Safari/537.36' } r = requests.get(url='https://wikipedia.org', headers=headers) print(i, r.status_code) except Exception as exc: time.sleep(60) print(traceback.format_exc()) print('continue') continue print('break') break print('Finished')

What do I need to change to get the desired behaviour?

Which loop should be broken when the request is successful? while or for? — K.Mat
– K.Mat, Commented Apr 7, 2021 at 12:28
Is "break" being printed? What about "finished"? Either you don't expect the for loop to initiate another loop of the while loop, or there is an error being caught every time you make a request. — gmdev
– gmdev, Commented Apr 7, 2021 at 12:29
the loop should have 15 attempts at running successfully - at which ever iteration of the loop it is successful at, it should end. — gdogg371
– gdogg371, Commented Apr 7, 2021 at 12:30

andmed · Accepted Answer · 2021-04-07 12:29:58Z

Remove while cycle. while work under for. When python come to break it "break" while cycle.

It's work for me:

import traceback import requests import time for i in range(0, 15): try: headers ={ 'authority': 'www.wikipedia.org', 'method': 'GET', 'path': '/wikipedia.org', 'scheme': 'https', 'accept': '*/*', 'accept-language': 'en-GB,en-US;q=0.9,en;q=0.8', 'referer': 'https://google.com', 'sec-fetch-dest': 'empty', 'sec-fetch-mode': 'cors', 'sec-fetch-site': 'same-origin', 'user-agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/87.0.4280.66 Safari/537.36' } r = requests.get(url='https://wikipedia.org', headers=headers) print(i, r.status_code) except Exception as exc: time.sleep(60) print(traceback.format_exc()) print('continue') continue print('break') break print('Finished')

Gignac · Accepted Answer · 2021-04-07 12:33:16Z

try this:

import requests from requests.exceptions import HTTPError def get_response(url): for i in range(15): try: response = requests.get(url) response.raise_for_status() return response.content except HTTPError as http_err: print(f"HTTP error occurred: {http_err}") except Exception as err: print(f"Other error occurred: {err}") url = 'http://google.com' response = get_response(url) print(response) url = 'http://no_valid_address.com' response = get_response(url) print(response)

Tom Wojcik · Accepted Answer · 2021-04-07 12:44:54Z

Expected behaviour is that if a request has been successful, the loop breaks and the code stops

the loop should have 15 attempts at running successfully - at which ever iteration of the loop it is successful at, it should end.

You don't need the while block. If you want to have 15 attempts or until first successful request, just do it.

import random def random_error(): if random.random() > 0.2: raise Exception for _ in range(15): print('-' * 20) try: random_error() except Exception as exc: print('exception raised') else: print('exception not raised') break print('Finished')

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