处理 urllib2 的超时?- Python [英] Handling urllib2's timeout? - Python

问题描述

我在 urllib2 的 urlopen 中使用超时参数.

I'm using the timeout parameter within the urllib2's urlopen.

urllib2.urlopen('http://www.example.org', timeout=1)

如何告诉 Python 如果超时到期，应该引发自定义错误?

How do I tell Python that if the timeout expires a custom error should be raised?

有什么想法吗?

推荐答案

在极少数情况下您要使用 ，除了:.这样做会捕获任何异常，这可能很难调试，并且它会捕获包括 SystemExit 和 KeyboardInterupt 在内的异常，这会使您的程序烦人使用..

There are very few cases where you want to use except:. Doing this captures any exception, which can be hard to debug, and it captures exceptions including SystemExit and KeyboardInterupt, which can make your program annoying to use..

最简单的情况是，您会发现 urllib2.URLError:

At the very simplest, you would catch urllib2.URLError:

try:
    urllib2.urlopen("http://example.com", timeout = 1)
except urllib2.URLError, e:
    raise MyException("There was an error: %r" % e)

以下应捕获连接超时时引发的特定错误:

The following should capture the specific error raised when the connection times out:

import urllib2
import socket

class MyException(Exception):
    pass

try:
    urllib2.urlopen("http://example.com", timeout = 1)
except urllib2.URLError, e:
    # For Python 2.6
    if isinstance(e.reason, socket.timeout):
        raise MyException("There was an error: %r" % e)
    else:
        # reraise the original error
        raise
except socket.timeout, e:
    # For Python 2.7
    raise MyException("There was an error: %r" % e)

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