如何重试 x 次并打印长时间睡眠

Question

我一直在试图了解退避是如何工作的。 我的目标是每当我达到 status_code 等：405 5 次。 我想休眠 60000 秒并打印出出现状态错误 405。

现在我已经写了：

import time

import backoff
import requests

@backoff.on_exception(
    backoff.expo,
    requests.exceptions.RequestException,
    max_tries=5,
    giveup=lambda e: e.response is not None and e.response.status_code == 405
)
def publish(url):
    r = requests.post(url, timeout=10)
    r.raise_for_status()


publish("https://www.google.se/")

现在发生的情况是，如果它只达到 405 一次，它将引发 status_code 并停止脚本。 我正在寻找的是如何让脚本重试 5 次，如果状态是 405 连续 5 次，那么我们想要长时间睡眠并打印出来。 如何使用 backofF 做到这一点？ 我也想提出其他建议：）

旧的计数器方式：

    import requests
    import time
    from requests.exceptions import ConnectionError, ReadTimeout, RequestException, Timeout
    
    exception_counter = 0
    
    while True:
    
        try:
            response = requests.get("https://stackoverflow.com/", timeout=12)
    
            if response.ok:
                print("Very nice")
                time.sleep(60)
    
            else:
                print(
                    f'[Response -> {response.status_code}]'
                    f'[Response Url -> {response.url}]'
                )
                time.sleep(60)
    
                if response.status_code == 403:
                    if exception_counter >= 10:
                        print("Hit limitation of counter: Response [403]")
                        time.sleep(4294968)
    
                    exception_counter += 1
    
        
    except (ConnectionError) as err:
        print(err)
        time.sleep(random.randint(1, 3))
        
        if exception_counter >= 10:
            print(f"Hit limitation of coonnectionerror {err}")
            time.sleep(4294968)
            continue
        
        exception_counter += 1
        continue
        
    except (ReadTimeout, Timeout) as err:
        print(err)
        time.sleep(random.randint(1, 3))
        continue

    except RequestException as err:
        print(err)
        time.sleep(random.randint(1, 3))
        continue

    except Exception as err:
        print(err)
        time.sleep(random.randint(1, 3))
        
        if exception_counter >= 10:
            print(f"Hit limitation of Exception {err}")
            time.sleep(4294968)
            continue
        
        exception_counter += 1
        continue

Answer 1

您没有说在睡眠 60000 秒后要做什么，所以我将其设置为在四次尝试后睡眠，然后在正确失败之前进行最后一次（第五次）尝试。

您可以使用on_backoff处理程序添加您要求的自定义逻辑。

此外，我重新调整了您的giveup ，您可能以错误的方式使用了 boolean。

import time
import backoff
import requests


def backoff_hdlr(details):
    print("backoff_hdlr", details)
    if details["tries"] >= 4:
        print(f"sleeping")
        time.sleep(1)  # 60000


@backoff.on_exception(
    backoff.expo,
    requests.exceptions.RequestException,
    max_tries=5,
    giveup=lambda e: e.response.status_code != 405,
    on_backoff=backoff_hdlr,
)
def publish(url):
    print(f"called publish with url={url}")
    r = requests.post(url, timeout=10)
    r.raise_for_status()


publish("https://www.google.se/")

/Users/michael/.conda/envs/mip_typing/bin/python /Users/michael/git/mip_typing/scratch_2.py
called publish with url=https://www.google.se/
backoff_hdlr {'target': <function publish at 0x7fe45c626b80>, 'args': ('https://www.google.se/',), 'kwargs': {}, 'tries': 1, 'elapsed': 1.5e-05, 'wait': 0.8697943681459608}
called publish with url=https://www.google.se/
backoff_hdlr {'target': <function publish at 0x7fe45c626b80>, 'args': ('https://www.google.se/',), 'kwargs': {}, 'tries': 2, 'elapsed': 1.144912, 'wait': 1.5425500028676453}
called publish with url=https://www.google.se/
backoff_hdlr {'target': <function publish at 0x7fe45c626b80>, 'args': ('https://www.google.se/',), 'kwargs': {}, 'tries': 3, 'elapsed': 2.949183, 'wait': 0.2052666718206697}
called publish with url=https://www.google.se/
backoff_hdlr {'target': <function publish at 0x7fe45c626b80>, 'args': ('https://www.google.se/',), 'kwargs': {}, 'tries': 4, 'elapsed': 3.418447, 'wait': 5.113712077372433}
sleeping
called publish with url=https://www.google.se/
Traceback (most recent call last):
...