python - 不可能捕获 asyncio.TimeoutError 吗？

Question

我使用 asyncio 来获取 url，有时会超时，尽我所能，我无法使用以下代码捕获 asyncio.TimeoutError!

async def fetch(url, session):
    """Fetch a url, using specified ClientSession."""
    async with session.get(url) as response:
        # print(f"fetching {url}")
        try:
            resp = await response.read()
        except asyncio.TimeoutError:
            return {"results": f"timeout error on {url}"}

        if response.status != 200:
            return {"error": f"server returned {response.status}"}

        return str(resp, 'utf-8').rstrip()

这是堆栈跟踪。我该怎么做才能捕获此异常并记录它而不是退出我的程序？

 resource: {…}  
 severity:  "ERROR"  
 textPayload:  "Traceback (most recent call last):
  File "/env/local/lib/python3.7/site-packages/google/cloud/functions/worker.py", line 346, in run_http_function
    result = _function_handler.invoke_user_function(flask.request)
  File "/env/local/lib/python3.7/site-packages/google/cloud/functions/worker.py", line 217, in invoke_user_function
    return call_user_function(request_or_event)
  File "/env/local/lib/python3.7/site-packages/google/cloud/functions/worker.py", line 210, in call_user_function
    return self._user_function(request_or_event)
  File "/user_code/main.py", line 230, in gcf_update_all_featured_podcasts
    loop.run_until_complete(future)  # loop until done
  File "/opt/python3.7/lib/python3.7/asyncio/base_events.py", line 573, in run_until_complete
    return future.result()
  File "/user_code/main.py", line 201, in fetch_all
    _ = await asyncio.gather(*tasks)  # gather task responses
  File "/user_code/main.py", line 181, in fetch
    async with session.get(url) as response:
  File "/env/local/lib/python3.7/site-packages/aiohttp/client.py", line 1005, in __aenter__
    self._resp = await self._coro
  File "/env/local/lib/python3.7/site-packages/aiohttp/client.py", line 497, in _request
    await resp.start(conn)
  File "/env/local/lib/python3.7/site-packages/aiohttp/client_reqrep.py", line 857, in start
    self._continue = None
  File "/env/local/lib/python3.7/site-packages/aiohttp/helpers.py", line 585, in __exit__
    raise asyncio.TimeoutError from None
concurrent.futures._base.TimeoutError

Answer 1

在回溯中，您可以看到异常是从您尝试向 url 发出请求的行引发的，但您的 try block 位于下面一层。

尝试这样:

async def fetch(url, session):
"""Fetch a url, using specified ClientSession."""
    try:
         async with session.get(url) as response:
         # print(f"fetching {url}")
            resp = await response.read()
            if response.status != 200:
                return {"error": f"server returned {response.status}"}

            return str(resp, 'utf-8').rstrip()

    except asyncio.TimeoutError:
        return {"results": f"timeout error on {url}"}