python urllib2超时

Question

好吧，我在 google 和 stackoverflow 中搜索了这个答案，几个小时后没有看到一个正确的答案来做这个......

我在这里粘贴了 4 个假定的 python 工作脚本示例，用于为不存在的 url 设置默认超时，超时设置为套接字和/或超时参数。

没有人工作，所以永远不会触发超时。

有什么想法吗？

第一个例子:

import urllib2

try:                
    header_s = {"User-Agent":"Mozilla/5.0 (X11; U; Linux i686) Gecko/20071127 Firefox/2.0.0.11"}

    req = urllib2.Request("http://www.nonexistantdomainurl.com/notexist.php",headers = header_s)


    print urllib2.urlopen(req, None, 5.0).read()

except urllib2.URLError, e:
    print "Url Error: %r" % e

except Exception,e:
  print "Fallo de tipo ",e

else: 
    print "all ok!"

第二个例子:

import urllib2

try:
    response = urllib2.urlopen("http://www.nonexistantdomainurl.com/notexist.php", None, 2.5)
except urllib2.URLError, e:
    print "Oops, timed out?"

第三个例子:

from urllib2 import Request, urlopen, URLError, HTTPError
import base64


req = Request('http://www.nonexistantdomainurl.com/notexist.php')

try:
    response = urlopen(req,timeout=5.0)   

except HTTPError, e:
    print 'The server couldn\'t fulfill the request.'
    print 'Error code: ', e.code
except URLError, e:
    print 'We failed to reach a server.'
    print 'Reason: ', e.reason

第四个例子:

import urllib2
import socket


socket.setdefaulttimeout(5)

try:
    response = urllib2.urlopen("http://www.faluquito.com/equipo.php",timeout=5.0).read()   


except urllib2.URLError, e:
    print "Url Error: %r" % e

Answer 1

>>> import urllib2
>>> import time
>>> import contextlib
>>>
>>> def timeit():
...   s = time.time()
...   try:
...     yield
...   except urllib2.URLError:
...     pass
...   print 'took %.3f secs' % (time.time() - s)
...
>>> timeit = contextlib.contextmanager(timeit)
>>> with timeit():
...   r = urllib2.urlopen('http://loc:8080', None, 2)
...
took 2.002 secs
>>> with timeit():
...   r = urllib2.urlopen('http://loc:8080', None, 5)
...
took 5.003 secs