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python - gai error at/home [Errno -2] 名称或服务未知

转载 作者:太空宇宙 更新时间:2023-11-04 10:57:31 26 4
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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()

检查我的本地变量产量:

连接:“0x8baa4ac 处的 httplib.HTTPSConnection 实例”params: 'token=[some string]&apiKey=[some string]'

(我进行此调用的说明是:

使用 token 进行 auth_info API 调用:网址:https://rpxnow.com/api/v2/auth_info参数:

接口(interface) key [一些字符串] token 你在上面提取的 token 值)

但我收到主题行中提到的错误。为什么?

最佳答案

您误解了 httplib 的文档。实例化 HTTPSConnection 的参数只是主机名。然后将实际路径作为第二个参数传递给 request。所以:

connection = httplib.HTTPSConnection('rpxnow.com')
connection.request('POST', '/api/v2/auth_info', params)

关于python - gai error at/home [Errno -2] 名称或服务未知,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/8661730/

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