Question

我正在尝试捕获函数调用中抛出的特定类型的异常。我在try / except块中包含了函数调用，其中except块捕获了抛出的特定异常。我仍然得到该异常的系统失败堆栈跟踪，除非我还包含所有异常的常规catch。在包含该块并检查被捕获的异常的类型时，我看到它正在捕获我想要在第一个块中捕获的异常类型。不知道为什么会这样。

背景信息：使用webapp2和ndb处理谷歌应用引擎应用。文件函数有一个 init .py，可以从exceptions.py

中导入所有异常

模拟代码和结构

utils的/功能/ exceptions.py

"""
Custom exception types
"""

class InvalidParamsException(Exception):
def __init__(self, msg):
    self.msg = msg

def __str__(self):
    return repr(self.msg)

模型/ models.py

import os, sys
sys.path.append(os.path.join(os.path.dirname(__file__), ".."))
import utils.functions as func
<-->
class ModelClass(ndb.Model):

    @classmethod
    def new(cls):
         <-->
         raise func.InvalidParamsException("Invalid Params to function!")
         <-->

routes.py

import utils.functions as func
from models import ModelClass

class ModelClassHandler(webapp2.RequestHandler):
    def post(self):
        try:
            new_model = ModelClass.new()
        except func.InvalidParamsException as e:
            logging.debug("Caught the right Exception!!")
        except Exception as e:
            logging.debug(":(")
            logging.debug("EXCEPTION TYPE - %s"%str(type(e)))

如果我不包括第二个常规除了块，我得到的输出是：

Traceback (most recent call last):
File "/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/lib/webapp2-2.5.2/webapp2.py", line 1535, in __call__
  rv = self.handle_exception(request, response, e)
File "/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/lib/webapp2-2.5.2/webapp2.py", line 1529, in __call__
  rv = self.router.dispatch(request, response)
File "/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/lib/webapp2-2.5.2/webapp2.py", line 1278, in default_dispatcher
  return route.handler_adapter(request, response)
File "/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/lib/webapp2-2.5.2/webapp2.py", line 1102, in __call__
  return handler.dispatch()
File "/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/lib/webapp2-2.5.2/webapp2.py", line 572, in dispatch
  return self.handle_exception(e, self.app.debug)
File "/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/lib/webapp2-2.5.2/webapp2.py", line 570, in dispatch
  return method(*args, **kwargs)
File "{{my_path}}/routes.py", line 58, in post
  new_model = ModelClass.new()
File "{{my_path}}/models/models.py", line 559, in new
  raise func.InvalidParamsException("Invalid Params to function!")
InvalidParamsException: 'Invalid Params to function!'

如果我确实包含了第二个块，我会优雅地传递路由/函数，并在日志中看到它：

DEBUG    2016-03-25 01:01:03,221 routes.py:66] EXCEPTION TYPE - <class 'utils.functions.exceptions.InvalidParamsException'>

帮助/指导非常感谢!!

Answer 1

似乎Python引发了在当前命名空间中导入的异常。我最好的evicende就是回溯的最后一行调用引发的异常“InvalidParamsException”而不是“somemodule.InvalidParamsException”。

因此，我建议解决名称空间冲突，将异常明确地导入到“routes.py”的命名空间中：

from utils.functions.exceptions import InvalidParamsException

并通过其现在解析的命名空间名称捕获异常：

except InvalidParamsException as inv_param_exp: <...>

尝试/除了没有捕获特定类型的异常

1 个答案: