停止在我的例外，而不是库代码 [英] Stop at exception in my, not library code

问题描述

我正在使用Python库开发一个应用程序 urllib ，而且由于无法访问URL，有时候会出现异常。

I'm developing an app using a Python library urllib and it is sometimes rising exceptions due to not being able to access an URL.

但是，异常已经升级到标准库堆栈中的近6个级别：

However, the exception is raised almost 6 levels into the standard library stack:

/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>

我通常在 ipython3 中运行代码％pdb 魔术打开，以防万一有异常，我可以立即检查它。但是为此，我必须将堆栈6级别下降到我的代码。

I usually run the code in ipython3 with the %pdb magic turned on so in case there is an exception I can inspect it immediately. However for this I have to go down the stack 6 levels to get to my code.

可以直接使用我的应用程序崩溃指向我的代码吗？

Is it achievable that my app crashes pointing to my code directly?

推荐答案

我会去修改代码：

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

except Exception as e:
    raise RuntimeError(e)

这样：

• ％pdb将你移动到您的代码


• 原始异常作为次要异常的参数保留。

您还可以使用monkeypatch urllib.request.urlopen（） function：

You may also monkeypatch urllib.request.urlopen() function:

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)

任何时候在 urlibopen（） / code>在上下文管理器范围内调用

Any time you have an exception raised in urlibopen() call within the context manager scope:

with MonkeyPatchUrllib():
    #your code here

％pdb将仅将您的代码移动一级。

%pdb will move you only 1 level away from your code.

使用 sys.exc_info（）可以保留更详细的原始异常上下文（如其追溯）。

With sys.exc_info() it is possible to preserve a more verbose context of the original exception (like its traceback).

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