如何在异常超时后重试并从dict中删除？

Question

我想增加抓取内容的覆盖率并避免超时，但是我想如果页面出现超时错误，然后重试，但是如果第二轮超时增加超时计数器，则重试URL，最多重试3次，如果没有增加，则从dict中删除

async def _worker(self, i):
    while True:
        url = await self.fetching.get()
        if url is None:
            return

        #logger.info(f'Fetch worker {i} is fetching a URL: {url}')
        try:
            site = await self._fetch_rss(url)
            await self._processing(site)
        except asyncio.TimeoutError as err:
            while self.timeouts[url] < 1:
                logger.info("adding {}".format(url))
                site = await self.fetch(url)
                await self.process(site)
                self.timeouts[url] += 1

Answer 1

通常，我用于“尝试n次，如果我从未成功，则出错”的模式如下：

for _ in range(MAX_TRIES):
   try:
       # do something
       pass
   except Exception:
       # handle exception
       pass
   else:
       break
else: # A lesser-known syntax; else after for is only run if the for *wasn't* broken out of
    # We never broke, which means we never had a success, so handle the failure here
    pass