我有一个循环,开头为for I,范围为(0,100)。正常情况下,它可以正常运行,但有时由于网络条件而出现故障。目前,我已经设置它,以便在失败时,它将继续在except子句中(继续到I的下一个数字)。

我是否可以将相同的数字重新分配给i,并再次运行失败的循环迭代?


当前回答

我喜欢laurent-laporte的回答。下面是我的版本,它包装在一个类与静态方法和一些例子。我实现了重试计数作为另一种重试方式。还增加了kwargs。

from typing import List
import time


class Retry:
    @staticmethod
    def onerror_retry(exception, callback, retries: int = 0, timeout: float = 0, timedelta: float = 0,
                      errors: List = None, **kwargs):
        """

        @param exception: The exception to trigger retry handling with.
        @param callback: The function that will potentially fail with an exception
        @param retries: Optional total number of retries, regardless of timing if this threshold is met, the call will
                        raise the exception.
        @param timeout: Optional total amount of time to do retries after which the call will raise an exception
        @param timedelta: Optional amount of time to sleep in between calls
        @param errors: A list to receive all the exceptions that were caught.
        @param kwargs: An optional key value parameters to pass to the function to retry.
        """
        for retry in Retry.__onerror_retry(exception, callback, retries, timeout, timedelta, errors, **kwargs):
            if retry: retry(**kwargs)  # retry will be None when all retries fail.

    @staticmethod
    def __onerror_retry(exception, callback, retries: int = 0, timeout: float = 0, timedelta: float = 0,
                        errors: List = None, **kwargs):
        end_time = time.time() + timeout
        continues = 0
        while True:
            try:
                yield callback(**kwargs)
                break
            except exception as ex:
                print(ex)
                if errors:
                    errors.append(ex)

                continues += 1
                if 0 < retries < continues:
                    print('ran out of retries')
                    raise

                if timeout > 0 and time.time() > end_time:
                    print('ran out of time')
                    raise
                elif timedelta > 0:
                    time.sleep(timedelta)


err = 0

#
# sample dumb fail function
def fail_many_times(**kwargs):
    global err
    err += 1
    max_errors = kwargs.pop('max_errors', '') or 1
    if err < max_errors:
        raise ValueError("I made boo boo.")
    print("Successfully did something.")

#
# Example calls
try:
    #
    # retries with a parameter that overrides retries... just because
    Retry.onerror_retry(ValueError, fail_many_times, retries=5, max_errors=3)
    err = 0
    #
    # retries that run out of time, with 1 second sleep between retries.
    Retry.onerror_retry(ValueError, fail_many_times, timeout=5, timedelta=1, max_errors=30)
except Exception as err:
    print(err)

其他回答

尝试次数= 3 而尝试: 试一试: ... ... <状态好了> 打破 除了: 尝试- = 1 Else: #已执行的唯一break未被引发 < >失败状态

在Python装饰器库中也有类似的东西。

请记住,它不测试异常,而是测试返回值。它会重新尝试,直到被修饰的函数返回True。

稍微修改一下版本就可以了。

使用这个装饰器,您可以轻松地控制错误

class catch:
    def __init__(self, max=1, callback=None):
        self.max = max 
        self.callback = callback 
    
    def set_max(self, max):
        self.max = max
    
    def handler(self, *args, **kwargs):
        self.index = 0
        while self.index < self.max: 
            self.index += 1
            try:
                self.func(self, *args, **kwargs)
        
            except Exception as error:
                if callable(self.callback):
                    self.callback(self, error, args, kwargs)
                
    def __call__(self, func):
        self.func = func
        return self.handler

import time
def callback(cls, error, args, kwargs):
    print('func args', args, 'func kwargs', kwargs)
    print('error', repr(error), 'trying', cls.index)
    if cls.index == 2:
        cls.set_max(4)
    
    else:
        time.sleep(1)
    
    
@catch(max=2, callback=callback)  
def test(cls, ok, **kwargs):
    raise ValueError('ok')

test(1, message='hello')

以下是我对这个问题的看法。下面的重试功能支持以下特性:

当调用成功时返回被调用函数的值 如果尝试失败,则引发被调用函数的异常 尝试次数限制(0表示无限) 在尝试之间等待(线性或指数) 仅当异常是特定异常类型的实例时重试。 可选的尝试记录

import time

def retry(func, ex_type=Exception, limit=0, wait_ms=100, wait_increase_ratio=2, logger=None):
    attempt = 1
    while True:
        try:
            return func()
        except Exception as ex:
            if not isinstance(ex, ex_type):
                raise ex
            if 0 < limit <= attempt:
                if logger:
                    logger.warning("no more attempts")
                raise ex

            if logger:
                logger.error("failed execution attempt #%d", attempt, exc_info=ex)

            attempt += 1
            if logger:
                logger.info("waiting %d ms before attempt #%d", wait_ms, attempt)
            time.sleep(wait_ms / 1000)
            wait_ms *= wait_increase_ratio

用法:

def fail_randomly():
    y = random.randint(0, 10)
    if y < 10:
        y = 0
    return x / y


logger = logging.getLogger()
logger.setLevel(logging.INFO)
logger.addHandler(logging.StreamHandler(stream=sys.stdout))

logger.info("starting")
result = retry.retry(fail_randomly, ex_type=ZeroDivisionError, limit=20, logger=logger)
logger.info("result is: %s", result)

更多信息请看我的帖子。

我倾向于限制重试次数,这样如果某个特定项目出现问题,你就可以继续进行下一个项目,如下:

for i in range(100):
  for attempt in range(10):
    try:
      # do thing
    except:
      # perhaps reconnect, etc.
    else:
      break
  else:
    # we failed all the attempts - deal with the consequences.