如何在python 3.0中通过http下载文件,解决bug？

我有一个脚本,我想继续使用,但看起来我要么找到一些解决方法来解决Python 3中的错误,要么降级回2.6,因此不得不降级其他脚本......

希望有人在这里找到了解决方法.

问题在于,由于Python 3.0中有关字节和字符串的新变化,并非所有的库代码都经过了明显的测试.

我有一个从Web服务器下载页面的脚本.这个脚本在python 2.6中传递了一个用户名和密码作为url的一部分,但在Python 3.0中,这不再起作用了.

例如,这个:

import urllib.request;
url = "http://username:password@server/file";
urllib.request.urlretrieve(url, "temp.dat");

失败,出现此异常:

Traceback (most recent call last):
  File "C:\Temp\test.py", line 5, in 
    urllib.request.urlretrieve(url, "test.html");
  File "C:\Python30\lib\urllib\request.py", line 134, in urlretrieve
    return _urlopener.retrieve(url, filename, reporthook, data)
  File "C:\Python30\lib\urllib\request.py", line 1476, in retrieve
    fp = self.open(url, data)
  File "C:\Python30\lib\urllib\request.py", line 1444, in open
    return getattr(self, name)(url)
  File "C:\Python30\lib\urllib\request.py", line 1618, in open_http
    return self._open_generic_http(http.client.HTTPConnection, url, data)
  File "C:\Python30\lib\urllib\request.py", line 1576, in _open_generic_http
    auth = base64.b64encode(user_passwd).strip()
  File "C:\Python30\lib\base64.py", line 56, in b64encode
    raise TypeError("expected bytes, not %s" % s.__class__.__name__)
TypeError: expected bytes, not str

显然,base64编码现在需要字节并输出一个字符串,因此urlretrieve(或其中的一些代码)构建一个username:password字符串,并尝试对此进行base64编码以进行简单授权,失败.

如果我改为尝试使用urlopen,就像这样:

import urllib.request;
url = "http://username:password@server/file";
f = urllib.request.urlopen(url);
contents = f.read();

然后它失败并出现此异常:

Traceback (most recent call last):
  File "C:\Temp\test.py", line 5, in 
    f = urllib.request.urlopen(url);
  File "C:\Python30\lib\urllib\request.py", line 122, in urlopen
    return _opener.open(url, data, timeout)
  File "C:\Python30\lib\urllib\request.py", line 359, in open
    response = self._open(req, data)
  File "C:\Python30\lib\urllib\request.py", line 377, in _open
    '_open', req)
  File "C:\Python30\lib\urllib\request.py", line 337, in _call_chain
    result = func(*args)
  File "C:\Python30\lib\urllib\request.py", line 1082, in http_open
    return self.do_open(http.client.HTTPConnection, req)
  File "C:\Python30\lib\urllib\request.py", line 1051, in do_open
    h = http_class(host, timeout=req.timeout) # will parse host:port
  File "C:\Python30\lib\http\client.py", line 620, in __init__
    self._set_hostport(host, port)
  File "C:\Python30\lib\http\client.py", line 632, in _set_hostport
    raise InvalidURL("nonnumeric port: '%s'" % host[i+1:])
http.client.InvalidURL: nonnumeric port: 'password@server'

显然,这个"下一代url检索库"中的url解析不知道如何处理url中的用户名和密码.

我还有其他选择吗？

直接来自Py3k文档:http://docs.python.org/dev/py3k/library/urllib.request.html#examples

import urllib.request
# Create an OpenerDirector with support for Basic HTTP Authentication...
auth_handler = urllib.request.HTTPBasicAuthHandler()
auth_handler.add_password(realm='PDQ Application',
                          uri='https://mahler:8092/site-updates.py',
                          user='klem',
                          passwd='kadidd!ehopper')
opener = urllib.request.build_opener(auth_handler)
# ...and install it globally so it can be used with urlopen.
urllib.request.install_opener(opener)
urllib.request.urlopen('http://www.example.com/login.html')