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

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。

其他回答

使用递归

for i in range(100):
    def do():
        try:
            ## Network related scripts
        except SpecificException as ex:
            do()
    do() ## invoke do() whenever required inside this loop

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)

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}')