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 ...

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。

在for循环中执行while True，将try代码放入其中，只有当代码成功时才退出while循环。

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

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

我最近用我的python解决了这个问题，我很高兴与stackoverflow的访问者分享，如果需要请给予反馈。

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}')