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

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

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)

带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()

最清晰的方法是显式地设置i。例如:

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

    except MyException:
        continue

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}')