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

使用这个装饰器,您可以轻松地控制错误

class catch:
    def __init__(self, max=1, callback=None):
        self.max = max 
        self.callback = callback 
    
    def set_max(self, max):
        self.max = max
    
    def handler(self, *args, **kwargs):
        self.index = 0
        while self.index < self.max: 
            self.index += 1
            try:
                self.func(self, *args, **kwargs)
        
            except Exception as error:
                if callable(self.callback):
                    self.callback(self, error, args, kwargs)
                
    def __call__(self, func):
        self.func = func
        return self.handler

import time
def callback(cls, error, args, kwargs):
    print('func args', args, 'func kwargs', kwargs)
    print('error', repr(error), 'trying', cls.index)
    if cls.index == 2:
        cls.set_max(4)
    
    else:
        time.sleep(1)
    
    
@catch(max=2, callback=callback)  
def test(cls, ok, **kwargs):
    raise ValueError('ok')

test(1, message='hello')

其他回答

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)

我使用这个,它可以用于任何函数:

def run_with_retry(func: callable, max_retries: int = 3, wait_seconds: int = 2, **func_params):
num_retries = 1
while True:
    try:
        return func(*func_params.values())
    except Exception as e:
        if num_retries > max_retries:
            print('we have reached maximum errors and raising the exception')
            raise e
        else:
            print(f'{num_retries}/{max_retries}')
            print("Retrying error:", e)
            num_retries += 1
            sleep(wait_seconds)

像这样调用:

    def add(val1, val2):
        return val1 + val2

    run_with_retry(func=add, param1=10, param2=20)

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

以下是我对这个问题的看法。下面的重试功能支持以下特性:

当调用成功时返回被调用函数的值 如果尝试失败,则引发被调用函数的异常 尝试次数限制(0表示无限) 在尝试之间等待(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)

更多信息请看我的帖子。