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

我喜欢使用bool值,如下所示:

success = False
num_try = 0
while success is False:
    if num_try >= 10: # or any number
        # handle error how  you please
    try:
        # code
        success = True
    except Exception as e:
        # record or do something with exception if needed
        num_try += 1

其他回答

我在我的代码中使用following,

   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

如果你想要一个没有嵌套循环和成功调用break的解决方案,你可以为任何可迭代对象开发一个快速的可检索包装。这里有一个我经常遇到的网络问题的例子——保存的身份验证过期。它的用法是这样的:

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

如果您正在寻找的是重新尝试x次失败的尝试,那么单个for else循环可能就是您想要的。考虑这个例子,尝试了3次:

attempts = 3

for attempt in range(1, attempts+1):
    try:
        if attempt < 4:
            raise TypeError(f"Error raised on attempt: {attempt}")
        else:
            print(f'Attempt {attempt} finally worked.')
    except (TypeError) as error:
        print(f'Attempt {attempt} hit the exception.')
        continue
    else:
        break
else:
    print(f'Exit after final attempt: {attempt}')

print(f'\nGo on to execute other code ...')

给出输出:

Attempt 1 hit the exception.
Attempt 2 hit the exception.
Attempt 3 hit the exception.
Exit after final attempt: 3

Go on to execute other code ...

再试一次它就成功了

attempts = 4

给出输出:

Attempt 1 hit the exception.
Attempt 2 hit the exception.
Attempt 3 hit the exception.
Attempt 4 finally worked.

Go on to execute other code ...

我使用这个,它可以用于任何函数:

def run_with_retry(func: callable, max_retries: int = 3, wait_seconds: int = 2, **func_params):
num_retries = 1
while True:
    try:
        return func(*func_params.values())
    except Exception as e:
        if num_retries > max_retries:
            print('we have reached maximum errors and raising the exception')
            raise e
        else:
            print(f'{num_retries}/{max_retries}')
            print("Retrying error:", e)
            num_retries += 1
            sleep(wait_seconds)

像这样调用:

    def add(val1, val2):
        return val1 + val2

    run_with_retry(func=add, param1=10, param2=20)

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)