我有一个我想继续使用的脚本,但看起来我要么不得不为Python3中的bug找到一些解决办法,要么降级回2.6,因此也不得不降级其他脚本……
希望这里有人已经找到了解决办法。
问题是,由于Python3.0中关于字节和字符串的新变化,显然并不是所有的库代码都经过了测试。
我有一个从web服务器下载页面的脚本。在python2.6中,此脚本将用户名和密码作为url的一部分进行传递,但在python3.0中,这不再起作用。
例如,如下所示:
import urllib.request;
url = "http://username:password@server/file";
urllib.request.urlretrieve(url, "temp.dat");失败,并出现此异常:
Traceback (most recent call last):
File "C:\Temp\test.py", line 5, in <module>
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编码现在需要字节数并输出一个字符串,因此构建username:password字符串并试图对其进行Base64编码以进行简单授权的urlretrieve (或其中的一些代码)失败。
如果我尝试使用urlopen,如下所示:
import urllib.request;
url = "http://username:password@server/file";
f = urllib.request.urlopen(url);
contents = f.read();然后它会失败,并出现以下异常:
Traceback (most recent call last):
File "C:\Temp\test.py", line 5, in <module>
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解析并不知道如何处理url中的用户名和密码。
我还有其他选择吗?
发布于 2008-12-27 22:04:53
直接从Py3k文档获取: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')发布于 2008-12-28 01:21:20
我的建议是将2.*分支保留为生产分支,直到您可以对3.0版本进行排序。
在迁移到Python3.0之前,我将等待一段时间。似乎有很多人在赶时间,但我只想把一切都理顺,并选择一个像样的第三方库。这可能需要一年,可能需要18个月,但“升级”的压力对我来说真的很低。
发布于 2021-03-25 17:51:47
您可以使用requests.get下载文件。尝试下面的示例代码:
import requests
from requests.auth import HTTPBasicAuth
def download_file(user_name, user_pwd, url, file_path):
file_name = url.rsplit('/', 1)[-1]
with requests.get(url, stream = True, auth = HTTPBasicAuth(user_name, user_pwd)) as response:
with open(file_path + "/" + file_name, 'wb') as f:
for chunk in response.iter_content(chunk_size = 8192):
f.write(chunk)
# You will download the login.html file to /home/dan/
download_file("dan", "password", "http://www.example.com/login.html", "/home/dan/")好好享受吧!!
https://stackoverflow.com/questions/395451
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