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

你可以使用Python重试包。 重试

它是用Python编写的，目的是简化向几乎任何东西添加重试行为的任务。

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

Decorator是一个很好的方法。

from functools import wraps
import time

class retry:
    def __init__(self, success=lambda r:True, times=3, delay=1, raiseexception=True, echo=True):
        self.success = success
        self.times = times
        self.raiseexception = raiseexception
        self.echo = echo
        self.delay = delay
    def retry(fun, *args, success=lambda r:True, times=3, delay=1, raiseexception=True, echo=True, **kwargs):
        ex = Exception(f"{fun} failed.")
        r = None
        for i in range(times):
            if i > 0:
                time.sleep(delay*2**(i-1))
            try:
                r = fun(*args, **kwargs)
                s = success(r)
            except Exception as e:
                s = False
                ex = e
                # raise e
            if not s:
                continue
            return r
        else:
            if echo:
                print(f"{fun} failed.", "args:", args, kwargs, "\nresult: %s"%r)
            if raiseexception:
                raise ex
    def __call__(self, fun):
        @wraps(fun)
        def wraper(*args, retry=0, **kwargs):
            retry = retry if retry>0 else self.times
            return self.__class__.retry(fun, *args, 
                                        success=self.success, 
                                        times=retry,
                                        delay=self.delay,
                                        raiseexception = self.raiseexception,
                                        echo = self.echo,
                                        **kwargs)
        return wraper

一些用法示例:

@retry(success=lambda x:x>3, times=4, delay=0.1)
def rf1(x=[]):
    x.append(1)
    print(x)
    return len(x)
> rf1()

[1]
[1, 1]
[1, 1, 1]
[1, 1, 1, 1]

4

@retry(success=lambda x:x>3, times=4, delay=0.1)
def rf2(l=[], v=1):
    l.append(v)
    print(l)
    assert len(l)>4
    return len(l)
> rf2(v=2, retry=10) #overwite times=4

[2]
[2, 2]
[2, 2, 2]
[2, 2, 2, 2]
[2, 2, 2, 2, 2]

5

> retry.retry(lambda a,b:a+b, 1, 2, times=2)

3

> retry.retry(lambda a,b:a+b, 1, "2", times=2)

TypeError: unsupported operand type(s) for +: 'int' and 'str'

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}')

以下是我关于如何解决这个问题的想法:

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)