在python中检查超时错误

问题描述

因此，在请求后，我有一个非常通用的日志记录语句：

So I have a pretty generic logging statement after a request:

try:
    r = requests.get(testUrl, timeout=10.0)
except Exception, err:
    logger.error({"message": err.message})

除了 TimeoutError 以外，它对我抛出的所有内容都非常有用。当请求超时时，我得到的是一个元组，它尝试并且无法序列化。

This works great for everything I've thrown at it except TimeoutError. When the request times out the err I get back is a tuple that it tries and fails to serialize.

我的问题是如何只捕获这种错误？对于初学者来说，我无法访问 TimeoutError 。我试过从例外导入中添加 * ，但是没有运气。我也尝试导入 OSError ，因为文档说 TimeoutError 是一个子类，但是我无法访问<$导入 OSError 后c $ c> TimeoutError 。

My question is how do I catch just this one type of error? For starters TimeoutError is not something I have access to. I have tried adding from exceptions import * but with no luck. I've also tried importing OSError because the docs say TimeoutError is a subclass, but I was unable to access TimeoutError after importing OSError.

我打算依次列出我的例外情况：

I plan to either list my exceptions in order:

except TimeoutError, err:
     #handle this specific error
except Exception, err:
     #handle all other errors

或仅检查类型：

except Exception, err:
    if isinstance(err, TimeoutError):
        #handle specific error
    #handle all other errors

Python 2.7.3& Django 1.5

Python 2.7.3 & Django 1.5

推荐答案

您可以处理例外：

You can handle requests.Timeout exception:

try:
    r = requests.get(testUrl, timeout=10.0)
except requests.Timeout as err:
    logger.error({"message": err.message})
except requests.RequestException as err:
    # handle other errors

示例：

>>> import requests
>>> url = "http://httpbin.org/delay/2"
>>> try:
...     r = requests.get(url, timeout=1)
... except requests.Timeout as err:
...     print(err.message)
...
HTTPConnectionPool(host='httpbin.org', port=80): Read timed out. (read timeout=1)

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