Handling urllib2's timeout? - Python

问题

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

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

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

Any ideas?

回答1:

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..

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)

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)

回答2:

In Python 2.7.3:

import urllib2
import socket

class MyException(Exception):
    pass

try:
    urllib2.urlopen("http://example.com", timeout = 1)
except urllib2.URLError as e:
    print type(e)    #not catch
except socket.timeout as e:
    print type(e)    #catched
    raise MyException("There was an error: %r" % e)

来源：https://stackoverflow.com/questions/2712524/handling-urllib2s-timeout-python

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