处理urllib2的超时？ - Python

Question

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?

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Any ideas?

Answer 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)

Answer 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)