问题：异常后如何重试？
我有一个以开头的循环for i in range(0, 100)。正常情况下，它可以正常运行，但有时由于网络条件而失败。目前，我已对其进行了设置，以便在失败时，它将continue在except子句中（继续到的下一个数字i）。
我是否可以将相同的数字重新分配给i循环并再次执行失败的循环？
回答 0
做一个while True内部的for循环，把你的try代码中，并突破从while只有当你的代码的成功循环。
for i in range(0,100):
while True:
try:
# do stuff
except SomeSpecificException:
continue
break回答 1
我更喜欢限制重试的次数，这样，如果该特定项目有问题，您最终将继续进行下一个，因此：
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.回答 2
在重试包是重试的代码块失败的好办法。
例如：
@retry(wait_random_min=1000, wait_random_max=2000)
def wait_random_1_to_2_s():
print("Randomly wait 1 to 2 seconds between retries")回答 3
这是一个与其他解决方案类似的解决方案，但是如果未按规定的次数或重试次数失败，则会引发异常。
tries = 3
for i in range(tries):
try:
do_the_thing()
except KeyError as e:
if i < tries - 1: # i is zero indexed
continue
else:
raise
break回答 4
没有使用那些难看的while循环的更“实用”的方法：
def tryAgain(retries=0):
if retries > 10: return
try:
# Do stuff
except:
retries+=1
tryAgain(retries)
tryAgain()回答 5
最清晰的方法是显式设置i。例如：
i = 0
while i < 100:
i += 1
try:
# do stuff
except MyException:
continue回答 6
带有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()回答 7
for _ in range(5):
try:
# replace this with something that may fail
raise ValueError("foo")
# replace Exception with a more specific exception
except Exception as e:
err = e
continue
# no exception, continue remainder of code
else:
break
# did not break the for loop, therefore all attempts
# raised an exception
else:
raise err我的版本与上述几种类似，但是没有使用单独的while循环，并且如果所有重试均失败，则会重新引发最新的异常。可以err = None在顶部显式设置，但不是严格必需的，因为只有在else出现错误并因此err被设置时才执行最后一个块。
回答 8
回答 9
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回答 10
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回答 11
您可以使用Python重试包。 重试
它是用Python编写的，以简化将重试行为添加到几乎所有内容的任务。
回答 12
retrying：tenacity和backoff（2020更新）的替代方案
将重试库是以前的路要走，但可悲的是它有一些缺陷，自2016年其他选择似乎还没有得到任何更新补偿和坚韧。在撰写本文时，坚韧程度更高的GItHub星星（2.3k和1.2k）并已更新，因此我选择使用它。这是一个例子：
from functools import partial
import random # producing random errors for this example
from tenacity import retry, stop_after_delay, wait_fixed, retry_if_exception_type
# Custom error type for this example
class CommunicationError(Exception):
pass
# Define shorthand decorator for the used settings.
retry_on_communication_error = partial(
retry,
stop=stop_after_delay(10), # max. 10 seconds wait.
wait=wait_fixed(0.4), # wait 400ms
retry=retry_if_exception_type(CommunicationError),
)()
@retry_on_communication_error
def do_something_unreliable(i):
if random.randint(1, 5) == 3:
print('Run#', i, 'Error occured. Retrying.')
raise CommunicationError()上面的代码输出类似：
Run# 3 Error occured. Retrying.
Run# 5 Error occured. Retrying.
Run# 6 Error occured. Retrying.
Run# 6 Error occured. Retrying.
Run# 10 Error occured. Retrying.
.
.
.tenacity.retry在韧度GitHub页面上列出了的更多设置。
回答 13
如果您想要一个没有嵌套循环且无需break成功调用的解决方案，则可retriable以为任何可迭代的对象开发一个快速包装。这是我经常遇到的网络问题的示例-保存的身份验证过期。它的使用将如下所示：
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回答 14
我在代码中使用以下代码，
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回答 15
attempts = 3
while attempts:
try:
...
...
<status ok>
break
except:
attempts -=1
else: # executed only break was not raised
<status failed>回答 16
这是我对这个问题的看法。以下retry功能支持以下功能：
- 成功时返回被调用函数的值
- 如果尝试已用尽，则引发被调用函数的异常
- 尝试次数的限制（0表示无限制）
- 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才重试。
- 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)有关更多信息，请参见我的帖子。
回答 17
这是我的解决方法：
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)回答 18
我最近与我的python合作解决了这个问题，很高兴与stackoverflow访问者分享它，如果需要的话请提供反馈。
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}')回答 19
仅在try子句成功时增加循环变量
