使用Python在Twitter 1.1中发布仅应用程序请求

2023年1月14日 139次阅读

我想使用仅应用程序身份验证访问Twitter 1.1搜索端点.为了做到这一点,我正在尝试实现Twitter API文档中给出的步骤 –
https://dev.twitter.com/docs/auth/application-only-auth(滚动到“发布仅应用程序请求”)

我无法在步骤2中获得“持票人令牌”.当我运行以下代码时,我收到“响应：302 Found”,这是一个重定向到位置：https://api.twitter.com/oauth2/token
理想情况下它应该是“200 OK”

import urllib
import base64
import httplib

CONSUMER_KEY = 'my_key'
CONSUMER_SECRET = 'my_secret'

encoded_CONSUMER_KEY = urllib.quote(CONSUMER_KEY)
encoded_CONSUMER_SECRET = urllib.quote(CONSUMER_SECRET)

concat_consumer_url = encoded_CONSUMER_KEY + ":" + encoded_CONSUMER_SECRET

host = 'api.twitter.com'
url = '/oauth2/token'
params = urllib.urlencode({'grant_type' : 'client_credentials'})
req = httplib.HTTP(host)
req.putrequest("POST", url)
req.putheader("Host", host)
req.putheader("User-Agent", "My Twitter 1.1")
req.putheader("Authorization", "Basic %s" % base64.b64encode(concat_consumer_url))
req.putheader("Content-Type" ,"application/x-www-form-urlencoded;charset=UTF-8")
req.putheader("Content-Length", "29")
req.putheader("Accept-Encoding", "gzip")

req.endheaders()
req.send(params)

# get the response
statuscode, statusmessage, header = req.getreply()
print "Response: ", statuscode, statusmessage
print "Headers: ", header

我不想使用任何Twitter API包装器来访问它.

最佳答案 问题是必须使用HTTPS连接调用URL.请检查有效的修改代码.

import urllib
import base64
import httplib

CONSUMER_KEY = 'my_key'
CONSUMER_SECRET = 'my_secret'

encoded_CONSUMER_KEY = urllib.quote(CONSUMER_KEY)
encoded_CONSUMER_SECRET = urllib.quote(CONSUMER_SECRET)

concat_consumer_url = encoded_CONSUMER_KEY + ":" + encoded_CONSUMER_SECRET

host = 'api.twitter.com'
url = '/oauth2/token/'
params = urllib.urlencode({'grant_type' : 'client_credentials'})
req = httplib.HTTPSConnection(host)
req.putrequest("POST", url)
req.putheader("Host", host)
req.putheader("User-Agent", "My Twitter 1.1")
req.putheader("Authorization", "Basic %s" % base64.b64encode(concat_consumer_url))
req.putheader("Content-Type" ,"application/x-www-form-urlencoded;charset=UTF-8")
req.putheader("Content-Length", "29")
req.putheader("Accept-Encoding", "gzip")

req.endheaders()
req.send(params)

resp = req.getresponse()
print resp.status, resp.reason
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