如何捕获 urllib.error.HTTPError: HTTP Error 403: Forbidden？

Question

我正在编写一个 Flask 应用程序，我在其中调用了一个可能返回 403 的方法。我认为以下代码可以适当地处理该错误：

try:
    connection = myLib.Login(username, password)
except urllib.error.HTTPError as err:
    abort(err.code)

但这似乎不起作用。如果 myLib.Login 返回 403，我会得到以下信息：

[2019-05-22 15:22:15,798] ERROR in app: Exception on /api/users [POST]
Traceback (most recent call last):
  File "/usr/local/lib/python3.7/site-packages/flask/app.py", line 2292, in wsgi_app
    response = self.full_dispatch_request()
  File "/usr/local/lib/python3.7/site-packages/flask/app.py", line 1815, in full_dispatch_request
    rv = self.handle_user_exception(e)
  File "/usr/local/lib/python3.7/site-packages/flask/app.py", line 1718, in handle_user_exception
    reraise(exc_type, exc_value, tb)
  File "/usr/local/lib/python3.7/site-packages/flask/_compat.py", line 35, in reraise
    raise value
  File "/usr/local/lib/python3.7/site-packages/flask/app.py", line 1813, in full_dispatch_request
    rv = self.dispatch_request()
  File "/usr/local/lib/python3.7/site-packages/flask/app.py", line 1799, in dispatch_request
    return self.view_functions[rule.endpoint](**req.view_args)
  File "./app/routes.py", line 16, in authenticate
    connection = myLib.Login(username, password)

  File "/usr/local/lib/python3.7/site-packages/mylib.py", line 90, in __open
    resp = opener.open(req)
  File "/usr/local/lib/python3.7/urllib/request.py", line 531, in open
    response = meth(req, response)
  File "/usr/local/lib/python3.7/urllib/request.py", line 641, in http_response
    'http', request, response, code, msg, hdrs)
  File "/usr/local/lib/python3.7/urllib/request.py", line 569, in error
    return self._call_chain(*args)
  File "/usr/local/lib/python3.7/urllib/request.py", line 503, in _call_chain
    result = func(*args)
  File "/usr/local/lib/python3.7/urllib/request.py", line 649, in http_error_default
    raise HTTPError(req.full_url, code, msg, hdrs, fp)
urllib.error.HTTPError: HTTP Error 403: Forbidden

如何捕获此错误并返回 403？

在常规 Python 脚本中，异常触发很好，所以我怀疑这可能与 flask 和/或 uwsgi 有关。

Answer 1

这很可能是一个问题

connection = myLib.Login(username, password)

是异步的。 myLib.Login 可能会产生一个子线程。由于他们有自己的上下文，父母无法捕捉到孩子提出的异常。 更多信息请参考Catch a thread's exception in the caller thread in Python。

编辑：

如果你想等待（“阻塞”）直到连接建立后再决定是继续还是中止，你可以使用queues来建立线程间的通信：

import queue
notifierQueue = queue.Queue()
connection = myLib.Login(username, password, notifierQueue)

response = notifierQueue.get(block=True)
if response [some condition]: 
    abort(err.code)
else:
    do whatever

Queue.get() 将阻塞设置为 true（默认）将等待队列中的项目可用。处理 HTTPError 需要在 Login 部分然后 put() 中进行。您还可以为 get() 设置超时参数，如果在时间结束后队列中没有元素，则会引发异常。

Answer 2

异常按预期工作。我的 devops 链是问题所在。