Python - SSLV3_ALERT_HANDSHAKE_FAILURE，这是密码问题还是证书问题？

Question

我正在尝试通过 python 请求 API 执行 https url，但是我收到错误:

  File "<console>", line 5, in <module>
  File "/usr/lib/python2.7/site-packages/requests/api.py", line 72, in get
    return request('get', url, params=params, **kwargs)
  File "/usr/lib/python2.7/site-packages/requests/api.py", line 58, in request
    return session.request(method=method, url=url, **kwargs)
  File "/usr/lib/python2.7/site-packages/requests/sessions.py", line 513, in request
    resp = self.send(prep, **send_kwargs)
  File "/usr/lib/python2.7/site-packages/requests/sessions.py", line 623, in send
    r = adapter.send(request, **kwargs)
  File "/usr/lib/python2.7/site-packages/requests/adapters.py", line 514, in send
    raise SSLError(e, request=request)
SSLError: [SSL: SSLV3_ALERT_HANDSHAKE_FAILURE] sslv3 alert handshake failure (_ssl.c:579)

我认为这可能是一个不受信任的警告，所以我使用了

requests.packages.urllib3.disable_warnings()

但这并没有奏效，进一步搜索表明它可能与密码有关？

我的代码如下:

import requests
username="ncm"
password="*****"
headers={'content-type':'application/json'}
response = requests.get('https://10.55.244.2/level/15/exec/-/traceroute/10.50.0.1/source/vlan/5/CR',
                        verify=False,
                        auth=(username, password),
                        headers=headers,
                        timeout=4)

证书出示

Answer 1

经过一些搜索我发现使用 urlib2 是可行的，下面的代码可以正常运行

import urllib2, base64
request = urllib2.Request('https://yoururl')
base64string = base64.b64encode('%s:%s' % (username, password))
request.add_header("Authorization", "Basic %s" % base64string)   
result = urllib2.urlopen(request).read()