Python urllib2 基本认证问题

Question

更新：根据 Lee 的评论，我决定将我的代码压缩为一个非常简单的脚本并从命令行运行它：

import urllib2
import sys

username = sys.argv[1]
password = sys.argv[2]
url = sys.argv[3]
print("calling %s with %s:%s\n" % (url, username, password))

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

req = urllib2.Request(url)
f = urllib2.urlopen(req)
data = f.read()
print(data)

不幸的是，它仍然不会生成 Authorization 标头（每个 Wireshark）:(

我在通过 urllib2 发送基本 AUTH 时遇到问题。我看了一下this article，并按照示例进行操作。我的代码：

passman = urllib2.HTTPPasswordMgrWithDefaultRealm()
passman.add_password(None, "api.foursquare.com", username, password)
urllib2.install_opener(urllib2.build_opener(urllib2.HTTPBasicAuthHandler(passman)))

req = urllib2.Request("http://api.foursquare.com/v1/user")    
f = urllib2.urlopen(req)
data = f.read()

我通过 wireshark 在 Wire 上看到以下内容：

GET /v1/user HTTP/1.1
Host: api.foursquare.com
Connection: close
Accept-Encoding: gzip
User-Agent: Python-urllib/2.5 

您可以看到授权未发送，而我通过 curl 发送请求时：curl -u user:password http://api.foursquare.com/v1/user

GET /v1/user HTTP/1.1
Authorization: Basic =SNIP=
User-Agent: curl/7.19.4 (universal-apple-darwin10.0) libcurl/7.19.4 OpenSSL/0.9.8k zlib/1.2.3
Host: api.foursquare.com
Accept: */*

由于某种原因，我的代码似乎没有发送身份验证 - 有人看到我遗漏了什么吗？

谢谢

-西蒙

Answer 1

问题可能是 Python 库，根据 HTTP 标准，首先发送未经身份验证的请求，然后只有当它通过 401 重试得到响应时，才会发送正确的凭据。如果 Foursquare 服务器不进行“完全标准的身份验证”，那么这些库将无法工作。

尝试使用标头进行身份验证：

import urllib2, base64

request = urllib2.Request("http://api.foursquare.com/v1/user")
base64string = base64.b64encode('%s:%s' % (username, password))
request.add_header("Authorization", "Basic %s" % base64string)   
result = urllib2.urlopen(request)

和你有同样的问题，从这个帖子找到了解决方案：http://forums.shopify.com/categories/9/posts/27662

Answer 2

这是我用来处理我在尝试访问 MailChimp 的 API 时遇到的类似问题的方法。这做同样的事情，只是格式更好。

import urllib2
import base64

chimpConfig = {
    "headers" : {
    "Content-Type": "application/json",
    "Authorization": "Basic " + base64.encodestring("hayden:MYSECRETAPIKEY").replace('\n', '')
    },
    "url": 'https://us12.api.mailchimp.com/3.0/'}

#perform authentication
datas = None
request = urllib2.Request(chimpConfig["url"], datas, chimpConfig["headers"])
result = urllib2.urlopen(request)

Answer 3

我建议当前的解决方案是使用我的包urllib2_prior_auth，它很好地解决了这个问题（我在inclusion 上工作到标准库。

Answer 4

（复制粘贴/改编自https://stackoverflow.com/a/24048772/1733117）。

首先，您可以继承urllib2.BaseHandler 或urllib2.HTTPBasicAuthHandler，并实现http_request，以便每个请求都有适当的Authorization 标头。

import urllib2
import base64

class PreemptiveBasicAuthHandler(urllib2.HTTPBasicAuthHandler):
    '''Preemptive basic auth.

    Instead of waiting for a 403 to then retry with the credentials,
    send the credentials if the url is handled by the password manager.
    Note: please use realm=None when calling add_password.'''
    def http_request(self, req):
        url = req.get_full_url()
        realm = None
        # this is very similar to the code from retry_http_basic_auth()
        # but returns a request object.
        user, pw = self.passwd.find_user_password(realm, url)
        if pw:
            raw = "%s:%s" % (user, pw)
            auth = 'Basic %s' % base64.b64encode(raw).strip()
            req.add_unredirected_header(self.auth_header, auth)
        return req

    https_request = http_request

那如果你和我一样懒，那就全局安装handler

api_url = "http://api.foursquare.com/"
api_username = "johndoe"
api_password = "some-cryptic-value"

auth_handler = PreemptiveBasicAuthHandler()
auth_handler.add_password(
    realm=None, # default realm.
    uri=api_url,
    user=api_username,
    passwd=api_password)
opener = urllib2.build_opener(auth_handler)
urllib2.install_opener(opener)

Answer 5

第二个参数必须是 URI，而不是域名。即

passman = urllib2.HTTPPasswordMgrWithDefaultRealm()
passman.add_password(None, "http://api.foursquare.com/", username, password)