在我的,而不是库代码中停止异常

Question

我正在使用Python库开发一个应用程序,urllib由于无法访问URL,有时会出现异常情况.

但是,异常会在标准库堆栈中引发近6个级别:

/home/user/Workspace/application/main.py in call(path)
     11                                  headers={'content-type': 'application/json'},
     12                                  data=b'')
---> 13     resp = urllib.request.urlopen(req)          ####### THIS IS MY CODE
     14     return json.loads(resp.read().decode('utf-8'))

/usr/lib/python3.4/urllib/request.py in urlopen(url, data, timeout, cafile, capath, cadefault, context)
    159     else:
    160         opener = _opener
--> 161     return opener.open(url, data, timeout)
    162 
    163 def install_opener(opener):

/usr/lib/python3.4/urllib/request.py in open(self, fullurl, data, timeout)
    461             req = meth(req)
    462 
--> 463         response = self._open(req, data)
    464 
    465         # post-process response

/usr/lib/python3.4/urllib/request.py in _open(self, req, data)
    479         protocol = req.type
    480         result = self._call_chain(self.handle_open, protocol, protocol +
--> 481                                   '_open', req)
    482         if result:
    483             return result

/usr/lib/python3.4/urllib/request.py in _call_chain(self, chain, kind, meth_name, *args)
    439         for handler in handlers:
    440             func = getattr(handler, meth_name)
--> 441             result = func(*args)
    442             if result is not None:
    443                 return result

/usr/lib/python3.4/urllib/request.py in http_open(self, req)
   1208 
   1209     def http_open(self, req):
-> 1210         return self.do_open(http.client.HTTPConnection, req)
   1211 
   1212     http_request = AbstractHTTPHandler.do_request_

/usr/lib/python3.4/urllib/request.py in do_open(self, http_class, req, **http_conn_args)
   1182                 h.request(req.get_method(), req.selector, req.data, headers)
   1183             except OSError as err: # timeout error
-> 1184                 raise URLError(err)
   1185             r = h.getresponse()
   1186         except:

URLError: <urlopen error [Errno 111] Connection refused>

我通常运行的代码在ipython3与%pdb魔术打开,以便在万一有例外,我可以马上检查它.然而,为此,我必须在堆栈6级别下达到我的代码.

我的应用程序直接崩溃指向我的代码是否可以实现？

Answer 1

我会去修改代码:

try:
    resp = urllib.request.urlopen(req)

except Exception as e:
    raise RuntimeError(e)

那样:

• %pdb将您带到您的代码,
• 原始异常将保留为"辅助"异常的参数.

你也可以monkeypatch urllib.request.urlopen()功能:

class MonkeyPatchUrllib(object):
    def __enter__(self):
        self.__urlopen = urllib.request.urlopen
        urllib.request.urlopen = self
    def __exit__(self, exception_type, exception_value, traceback):
        urllib.request.urlopen = self.__urlopen
    def __call__(self, *args, **kwargs):
        try:                                  
            return self.__urlopen(*args, **kwargs)
        except Exception as e:
            raise RuntimeError(e)

每次在urlibopen()上下文管理器范围内调用时都会引发异常:

with MonkeyPatchUrllib():
    #your code here

%pdb只会使您的代码离您只有1级.

[编辑]

使用sys.exc_info()它可以保留原始异常的更详细的上下文(如其回溯).