/home 出现 gai 错误 [Errno -2] 名称或服务未知
根据 httplib 文档中的示例：
>>> import httplib, urllib
>>> params = urllib.urlencode({'@number': 12524, '@type': 'issue', '@action': 'show'})
>>> headers = {"Content-type": "application/x-www-form-urlencoded",
... "Accept": "text/plain"}
>>> conn = httplib.HTTPConnection("bugs.python.org")
>>> conn.request("POST", "", params, headers)
>>> response = conn.getresponse()
>>> print response.status, response.reason
302 Found
>>> data = response.read()
>>> data
'Redirecting to <a href="http://bugs.python.org/issue12524">http://bugs.python.org/issue12524</a>'
>>> conn.close()
我的代码是：
import httplib
import urllib
token = request.POST.get('token')
if token:
params = urllib.urlencode({'apiKey':'[some string]', 'token':token})
connection = httplib.HTTPSConnection('rpxnow.com/api/v2/auth_info')
connection.request('POST', "", params)
response = connection.getresponse()
print response.read()
检查我的本地变量 yeilds：
连接：“httplib.HTTPSConnection 实例位于 0x8baa4ac” params: 'token=[some string]&apiKey=[some string]'
（我进行此调用的说明是：
使用令牌进行 auth_info API 调用： 网址：https://rpxnow.com/api/v2/auth_info 参数：
apiKey [一些字符串] 代币 您在上面提取的令牌值）
但我收到主题行中提到的错误。为什么？
per the example in the httplib docs:
>>> import httplib, urllib
>>> params = urllib.urlencode({'@number': 12524, '@type': 'issue', '@action': 'show'})
>>> headers = {"Content-type": "application/x-www-form-urlencoded",
... "Accept": "text/plain"}
>>> conn = httplib.HTTPConnection("bugs.python.org")
>>> conn.request("POST", "", params, headers)
>>> response = conn.getresponse()
>>> print response.status, response.reason
302 Found
>>> data = response.read()
>>> data
'Redirecting to <a href="http://bugs.python.org/issue12524">http://bugs.python.org/issue12524</a>'
>>> conn.close()
my code is:
import httplib
import urllib
token = request.POST.get('token')
if token:
params = urllib.urlencode({'apiKey':'[some string]', 'token':token})
connection = httplib.HTTPSConnection('rpxnow.com/api/v2/auth_info')
connection.request('POST', "", params)
response = connection.getresponse()
print response.read()
inspection of my local vars yeilds:
connection: "httplib.HTTPSConnection instance at 0x8baa4ac"
params: 'token=[some string]&apiKey=[some string]'
(My instructions to make this call are:
Use the token to make the auth_info API call:
URL: https://rpxnow.com/api/v2/auth_info
Parameters:
apiKey
[some string]
token
The token value you extracted above)
but I'm getting the error mentioned in the subject line. Why?
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评论（3）
我不知道 rpxnow.com 是什么，也不熟悉他们的 API，但此错误消息表明他们没有响应该 URL 上的请求的服务（即“rpxnow.com/api/v2/auth_info”） ）。
您能否验证他们的服务是否已在该 URL 上启动并运行？
I don't know what rpxnow.com is and I am not familiar with their API, but this error message indicates that they do not have a service responding to requests at that URL (i.e. 'rpxnow.com/api/v2/auth_info').
Are you able to verify that their service is up and running at that URL?
尝试使用这个：
http://docs. python-requests.org/en/latest/user/quickstart/#make-a-post-request
Try using this:
http://docs.python-requests.org/en/latest/user/quickstart/#make-a-post-request
您误解了 httplib 的文档。实例化
HTTPSConnection的参数只是主机名。然后，您将实际路径作为第二个参数传递给request。所以：You've misunderstood the documentation to httplib. The parameter to instantiate the
HTTPSConnectionis just the hostname. You then pass the actual path as the second param torequest. So: