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 ...

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

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}')

这里有一个快速装饰器来处理这个问题。7行，没有依赖关系。

def retry(exception=Exception, retries=3, delay=0):
    def wrap(func):
        for i in range(retries):
            try:
                return func()
            except exception as e:
                print(f'Retrying {func.__name__}: {i}/{retries}')
                time.sleep(delay)
        raise e
    return wrap

@retry()
def do_something():
  ...
@retry(HTTPError, retries=100, delay=3)
def download_something():
  ...

可以添加的一个功能是扩展异常以处理多个异常(splat一个列表)。

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.