0

我正在使用此stackoverflow 帖子中的代码来缩短 url...

import httplib
import urlparse

def unshorten_url(url):
    parsed = urlparse.urlparse(url)
    h = httplib.HTTPConnection(parsed.netloc)
    resource = parsed.path
    if parsed.query != "":
        resource += "?" + parsed.query
    h.request('HEAD', resource )
    response = h.getresponse()
    if response.status/100 == 3 and response.getheader('Location'):
        return unshorten_url(response.getheader('Location')) # changed to process chains of short urls
    else:
        return url

除新创建的 bit.ly 网址外，所有缩短的链接都不会缩短。

我收到此错误：

>>> unshorten_url("bit.ly/1atTViN")
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 7, in unshorten_url
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/httplib.py", line 955, in request
    self._send_request(method, url, body, headers)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/httplib.py", line 989, in _send_request
    self.endheaders(body)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/httplib.py", line 951, in endheaders
    self._send_output(message_body)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/httplib.py", line 811, in _send_output
    self.send(msg)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/httplib.py", line 773, in send
    self.connect()
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/httplib.py", line 754, in connect
    self.timeout, self.source_address)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/socket.py", line 571, in create_connection
    raise err
socket.error: [Errno 61] Connection refused

是什么赋予了？

4

2 回答 2

3

您忘记包含 URL 方案：

unshorten_url("http://bit.ly/1atTViN")

注意http://那里，这很重要。没有它，URL 将无法正确解析：

>>> import urlparse
>>> urlparse.urlparse('bit.ly/1atTViN')
ParseResult(scheme='', netloc='', path='bit.ly/1atTViN', params='', query='', fragment='')
>>> urlparse.urlparse('http://bit.ly/1atTViN')
ParseResult(scheme='http', netloc='bit.ly', path='/1atTViN', params='', query='', fragment='')

看看包含netlocno 时参数如何为空http://；您最终尝试连接到您自己的机器，并且您没有运行网络服务器，因此连接被拒绝。

于 2013-10-13T08:05:18.167 回答
0

可能 bit.ly 拒绝来自 httplib 等工具的连接。您可以尝试像这样更改用户代理：

h.putheader('User-Agent','Mozilla/5.0 (X11; U; Linux i686; pl-PL; rv:1.7.10) Gecko/20050717 Firefox/1.0.6')
于 2013-10-13T07:59:11.057 回答