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Python在不使用decorator的情况下使用坚韧性重试

python-2.7 python

tuk · 技术社区 · 7 年前

我正在尝试使用“坚韧”进行重试(没有装饰)。我的代码看起来就像解释的那样 here .

import logging
from tenacity import retry
import tenacity


def print_msg():
    try:
        logging.info('Hello')
        logging.info("World")
        raise Exception('Test error')
    except Exception as e:
        logging.error('caught error')
        raise e


if __name__ == '__main__':
    logging.basicConfig(
        format='%(asctime)s,%(msecs)d %(levelname)-8s [%(filename)s:%(lineno)d] %(message)s',
        datefmt='%d-%m-%Y:%H:%M:%S',
        level=logging.INFO)
    logging.info('Starting')
    try:
        r = tenacity.Retrying(
            tenacity.stop_after_attempt(2),
            tenacity.wait_incrementing(start=10, increment=100, max=1000),
            reraise=True
        )
        try:
            r.call(print_msg)
        except Exception:
            logging.error('Test error 2')
    except Exception:
        logging.error('Received Exception')

/Users/dmanna/PycharmProjects/demo/venv/bin/python /Users/dmanna/PycharmProjects/demo/retrier.py
25-11-2018:00:29:47,140 INFO     [retrier.py:21] Starting
25-11-2018:00:29:47,140 INFO     [retrier.py:8] Hello
25-11-2018:00:29:47,140 INFO     [retrier.py:9] World
25-11-2018:00:29:47,140 ERROR    [retrier.py:12] caught error
25-11-2018:00:29:47,141 ERROR    [retrier.py:31] Test error 2

Process finished with exit code 0

有人能告诉我出了什么问题,因为我在上面的代码中没有看到任何重试吗?

1 回复 | 直到 6 年前

tuk 7 年前

这个问题已经得到了回答 here . 交叉张贴答案

r = tenacity.Retrying(
            tenacity.stop_after_attempt(2),
            tenacity.wait_incrementing(start=10, increment=100, max=1000),
            reraise=True
        )

这意味着:

r = tenacity.Retrying(
            sleep=tenacity.stop_after_attempt(2),
            stop=tenacity.wait_incrementing(start=10, increment=100, max=1000),
            reraise=True
        )

你想要:

r = tenacity.Retrying(
            stop=tenacity.stop_after_attempt(2),
            wait=tenacity.wait_incrementing(start=10, increment=100, max=1000),
            reraise=True
        )

ingyhere 6 年前

注意:对于Python3.4+用户(支持注释),要明确地强制进行顽强的重试,一般的用例是使用简单的注释对其进行注释 @retry 然后提出特殊例外 TryAgain

@retry
def do_something():
    result = something_else()
    if result == 23:
       raise TryAgain

这种方法类似于Python2.7的答案 on this page 还可以实施:

import tenacity

def do_something():
    result = something_else()
    if result == 23:
       raise TryAgain

r = tenacity.Retrying(
        stop=tenacity.stop_after_attempt(2),
        wait=tenacity.wait_incrementing(start=10, increment=100, max=1000)
    )
r.wraps(do_something)

API documentation on whether to retry 或者 source that defines annotations