python urllib2基本身份验证

2024-03-29 13:51:19 发布

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嗨,我正试图使用python访问使用urllib2的API URL:

import urllib2

url = 'https://XXXXXXXXXX.com/'
username = 'XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX'
password = 'XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX'

passman = urllib2.HTTPPasswordMgrWithDefaultRealm()
passman.add_password(None, url, username, password)
authhandler = urllib2.HTTPBasicAuthHandler(passman)
opener = urllib2.build_opener(authhandler)
urllib2.install_opener(opener)
pagehandle = urllib2.urlopen(url)

我不知道领域是什么,但我假设我可以使用默认值,即无。

不管怎样,我还是有401错误:

Traceback (most recent call last): File "test5.py", line 12, in pagehandle = urllib2.urlopen(url) File "/usr/lib/python2.6/urllib2.py", line 126, in urlopen return _opener.open(url, data, timeout) File "/usr/lib/python2.6/urllib2.py", line 397, in open response = meth(req, response) File "/usr/lib/python2.6/urllib2.py", line 510, in http_response 'http', request, response, code, msg, hdrs) File "/usr/lib/python2.6/urllib2.py", line 435, in error return self._call_chain(*args) File "/usr/lib/python2.6/urllib2.py", line 369, in _call_chain result = func(*args) File "/usr/lib/python2.6/urllib2.py", line 518, in http_error_default raise HTTPError(req.get_full_url(), code, msg, hdrs, fp) urllib2.HTTPError: HTTP Error 401: Unauthorized

所以看来认证有问题。很可能使用realm=None吗?

谢谢!


Tags: inpyhttpurlresponselibusrline
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1楼 · 发布于 2024-03-29 13:51:19

我不明白为什么这个方法不起作用,但我成功地使用以下代码调用了API:

base64string = base64.encodestring('%s:%s' % (username, password)).replace('\n', '')

request = urllib2.Request(url)

request.add_header("Authorization", "Basic %s" % base64string) 

result = urllib2.urlopen(request)
        data = result.read()

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