[已解决]urllib2.URLError: <urlopen error [Errno 111] Connection refused>

niliuhe

按照python核心编程书上来的，例20.1HTTP认证客户端，但是还是报错，代码和报错信息分别如下，求大神解答！
代码：
#!/usr/bin/env python

import urllib2

LOGIN = 'wesc'
PASSWD = "you'll NeverGuess"
URL = 'http://localhost'

def handler_version(url):
    from urlparse import urlparse as up
    hdlr = urllib2.HTTPBasicAuthHandler()
    hdlr.add_password('Archives', up(url)[1], LOGIN, PASSWD)
    opener = urllib2.build_opener(hdlr)
    urllib2.install_opener(opener)
    return url

def request_version(url):
    from base64 import encodestring
    req = urllib2.Request(url)
    b64str = encodestring('%s:%s' % (LOGIN, PASSWD))[:-1]
    req.add_header("Authorization", "Basic %s" % b64str)
    return req

for funcType in ('handler', 'request'):
    print '*** Using %s:' % funcType.upper()
    url = eval('%s_version' % funcType)(URL)
    f = urllib2.urlopen(url)
    print f.readline()
    f.close()

执行后报错：
*** Using HANDLER:
Traceback (most recent call last):
  File "urlopenAuth.py", line 27, in <module>
    f = urllib2.urlopen(url)
  File "/usr/lib/python2.7/urllib2.py", line 154, in urlopen
    return opener.open(url, data, timeout)
  File "/usr/lib/python2.7/urllib2.py", line 429, in open
    response = self._open(req, data)
  File "/usr/lib/python2.7/urllib2.py", line 447, in _open
    '_open', req)
  File "/usr/lib/python2.7/urllib2.py", line 407, in _call_chain
    result = func(*args)
  File "/usr/lib/python2.7/urllib2.py", line 1228, in http_open
    return self.do_open(httplib.HTTPConnection, req)
  File "/usr/lib/python2.7/urllib2.py", line 1198, in do_open
    raise URLError(err)
urllib2.URLError: <urlopen error [Errno 111] Connection refused>

最佳答案

月排行榜 / 总排行榜

SixPy

2016-8-20 10:15:07

认真看书，按书里的步骤来~

跳转到最佳答案楼层

SixPy

URL = 'http://localhost'

你知道这句的含义吗？

niliuhe

SixPy 发表于 2016-8-20 09:59
URL = 'http://localhost'

你知道这句的含义吗？

新手不太懂，这不是表示本地主机么

SixPy

认真看书，按书里的步骤来~