How to wait for a specific http response in python

Question

I have a problem waiting for a response 200. When I perform my get request the expected resource isn't there and the response is 204. What I want to do is wait for two minutes and lets say call every 10 seconds to check is the response is 200. If time is out, I want to return an error.

def waitForResourceAvailable(response):
    while response.status_code == 204:
        time.sleep(10)
        if response.status_code == 200:
            break

This works well, but I need a solution to exit the loop in case of status code 200 not to appear in a specified time.

HTTP code 204 or 404 or even 50x - are absolutely normal response codes. Why do you expect 200 exactly? Are you responsible for server application? Which url are you trying? — Alex Yu
– Alex Yu, Commented Feb 11, 2019 at 15:48
response is fixed; neither it nor any part of it is going to change in the code you show. You have to repeat the request if you want a different response. — chepner
– chepner, Commented Feb 11, 2019 at 15:56
I run a command via REST-API and this command generates a report. When starting the command the response contains the reportID, but the report is not generated yet. So my next command to fetch the report has to wait until the resource is available. — Aurelio
– Aurelio, Commented Feb 11, 2019 at 15:57

Rolando cq · Accepted Answer · 2019-02-11 15:55:14Z

Be aware that if the value of the response changes while checking in the while condition it will also stop the iteration, so the if statement inside the while is not needed.

Note: the response object has to be able to change its value otherwise it will always be the same as it is a just a variable and not a function that checks a response

You can add a simple if statement with a time value

def waitForResourceAvailable(response, time):
    timer = 0
    while response.status_code == 204:
        time.sleep(10)
        timer += 10
        if timer > time:
            break
        if response.status_code == 200:
            break

Making it even better would be like:

def waitForResourceAvailable(response, timeout, timewait):
    timer = 0
    while response.status_code == 204:
        time.sleep(timewait)
        timer += timewait
        if timer > timeout:
            break
        if response.status_code == 200:
            break

aidanmelen · Accepted Answer · 2022-01-03 00:01:23Z

0

Check out the backoff library. You can annotate your functions such that when a requests fails; it will "backoff" and retry. Here a simple example below but it can be expanded for a lot more complex use cases such as: Give Up Conditions, Jitter, Using multiple decorators, etc.

@backoff.on_exception(backoff.expo,
                      (requests.exceptions.Timeout,
                       requests.exceptions.ConnectionError))
def get_url(url):
    return requests.get(url)

The on_exception decorator is used to retry when a specified exception is raised. Here’s an example using exponential backoff when any requests exception is raised:

answered Jan 3, 2022 at 0:01

aidanmelen

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