异常后如何重试？

我有一个以开头的循环for i in range(0, 100)。正常情况下，它可以正常运行，但有时由于网络条件而失败。目前，我已对其进行了设置，以便在失败时，它将continue在except子句中（继续到的下一个数字i）。

我是否可以将相同的数字重新分配给i循环并再次执行失败的循环？

做一个while True内部的for循环，把你的try代码中，并突破从while只有当你的代码的成功循环。

for i in range(0,100):
    while True:
        try:
            # do stuff
        except SomeSpecificException:
            continue
        break

我更喜欢限制重试的次数，这样，如果该特定项目有问题，您最终将继续进行下一个，因此：

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.

在重试包是重试的代码块失败的好办法。

例如：

@retry(wait_random_min=1000, wait_random_max=2000)
def wait_random_1_to_2_s():
    print("Randomly wait 1 to 2 seconds between retries")

这是一个与其他解决方案类似的解决方案，但是如果未按规定的次数或重试次数失败，则会引发异常。

tries = 3
for i in range(tries):
    try:
        do_the_thing()
    except KeyError as e:
        if i < tries - 1: # i is zero indexed
            continue
        else:
            raise
    break

没有使用那些难看的while循环的更“实用”的方法：

def tryAgain(retries=0):
    if retries > 10: return
    try:
        # Do stuff
    except:
        retries+=1
        tryAgain(retries)

tryAgain()

最清晰的方法是显式设置i。例如：

i = 0
while i < 100:
    i += 1
    try:
        # do stuff

    except MyException:
        continue

带有超时的通用解决方案：

import time

def onerror_retry(exception, callback, timeout=2, timedelta=.1):
    end_time = time.time() + timeout
    while True:
        try:
            yield callback()
            break
        except exception:
            if time.time() > end_time:
                raise
            elif timedelta > 0:
                time.sleep(timedelta)

用法：

for retry in onerror_retry(SomeSpecificException, do_stuff):
    retry()

for _ in range(5):
    try:
        # replace this with something that may fail
        raise ValueError("foo")

    # replace Exception with a more specific exception
    except Exception as e:
        err = e
        continue

    # no exception, continue remainder of code
    else:
        break

# did not break the for loop, therefore all attempts
# raised an exception
else:
    raise err

我的版本与上述几种类似，但是没有使用单独的while循环，并且如果所有重试均失败，则会重新引发最新的异常。可以err = None在顶部显式设置，但不是严格必需的，因为只有在else出现错误并因此err被设置时才执行最后一个块。

Python Decorator库中有类似的东西。

请记住，它不会测试异常，而是返回值。重试，直到修饰的函数返回True。

稍加修改的版本应该可以解决问题。

使用递归

for i in range(100):
    def do():
        try:
            ## Network related scripts
        except SpecificException as ex:
            do()
    do() ## invoke do() whenever required inside this loop

使用while和计数器：

count = 1
while count <= 3:  # try 3 times
    try:
        # do_the_logic()
        break
    except SomeSpecificException as e:
        # If trying 3rd time and still error?? 
        # Just throw the error- we don't have anything to hide :)
        if count == 3:
            raise
        count += 1

您可以使用Python重试包。 重试

它是用Python编写的，以简化将重试行为添加到几乎所有内容的任务。

retrying：tenacity和backoff（2020更新）的替代方案

将重试库是以前的路要走，但可悲的是它有一些缺陷，自2016年其他选择似乎还没有得到任何更新补偿和坚韧。在撰写本文时，坚韧程度更高的GItHub星星（2.3k和1.2k）并已更新，因此我选择使用它。这是一个例子：

from functools import partial
import random # producing random errors for this example

from tenacity import retry, stop_after_delay, wait_fixed, retry_if_exception_type

# Custom error type for this example
class CommunicationError(Exception):
    pass

# Define shorthand decorator for the used settings.
retry_on_communication_error = partial(
    retry,
    stop=stop_after_delay(10),  # max. 10 seconds wait.
    wait=wait_fixed(0.4),  # wait 400ms 
    retry=retry_if_exception_type(CommunicationError),
)()


@retry_on_communication_error
def do_something_unreliable(i):
    if random.randint(1, 5) == 3:
        print('Run#', i, 'Error occured. Retrying.')
        raise CommunicationError()

上面的代码输出类似：

Run# 3 Error occured. Retrying.
Run# 5 Error occured. Retrying.
Run# 6 Error occured. Retrying.
Run# 6 Error occured. Retrying.
Run# 10 Error occured. Retrying.
.
.
.

tenacity.retry在韧度GitHub页面上列出了的更多设置。

如果您想要一个没有嵌套循环且无需break成功调用的解决方案，则可retriable以为任何可迭代的对象开发一个快速包装。这是我经常遇到的网络问题的示例-保存的身份验证过期。它的使用将如下所示：

client = get_client()
smart_loop = retriable(list_of_values):

for value in smart_loop:
    try:
        client.do_something_with(value)
    except ClientAuthExpired:
        client = get_client()
        smart_loop.retry()
        continue
    except NetworkTimeout:
        smart_loop.retry()
        continue

我在代码中使用以下代码，

   for i in range(0, 10):
    try:
        #things I need to do
    except ValueError:
        print("Try #{} failed with ValueError: Sleeping for 2 secs before next try:".format(i))
        time.sleep(2)
        continue
    break

这是我对这个问题的看法。以下retry功能支持以下功能：

• 成功时返回被调用函数的值
• 如果尝试已用尽，则引发被调用函数的异常
• 尝试次数的限制（0表示无限制）
• 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才重试。
• 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)

有关更多信息，请参见我的帖子。

这是我的解决方法：

j = 19
def calc(y):
    global j
    try:
        j = j + 8 - y
        x = int(y/j)   # this will eventually raise DIV/0 when j=0
        print("i = ", str(y), " j = ", str(j), " x = ", str(x))
    except:
        j = j + 1   # when the exception happens, increment "j" and retry
        calc(y)
for i in range(50):
    calc(i)

我最近与我的python合作解决了这个问题，很高兴与stackoverflow访问者分享它，如果需要的话请提供反馈。

print("\nmonthly salary per day and year converter".title())
print('==' * 25)


def income_counter(day, salary, month):
    global result2, result, is_ready, result3
    result = salary / month
    result2 = result * day
    result3 = salary * 12
    is_ready = True
    return result, result2, result3, is_ready


i = 0
for i in range(5):
    try:
        month = int(input("\ntotal days of the current month: "))
        salary = int(input("total salary per month: "))
        day = int(input("Total Days to calculate> "))
        income_counter(day=day, salary=salary, month=month)
        if is_ready:
            print(f'Your Salary per one day is: {round(result)}')
            print(f'your income in {day} days will be: {round(result2)}')
            print(f'your total income in one year will be: {round(result3)}')
            break
        else:
            continue
    except ZeroDivisionError:
        is_ready = False
        i += 1
        print("a month does'nt have 0 days, please try again")
        print(f'total chances left: {5 - i}')
    except ValueError:
        is_ready = False
        i += 1
        print("Invalid value, please type a number")
        print(f'total chances left: {5 - i}')

仅在try子句成功时增加循环变量