[Instagram][Python][GoogleAppEngine]認証(ログイン)方法

Instagramの認証を読みながら実際に、ログイン方法について調べる。

Google App Engineを使った場合は、下のように実装できるようだ。

#!/usr/bin/env python
# -*- coding: utf-8 -*-
#
# Copyright 2007 Google Inc.
#
# Licensed under the Apache License, Version 2.0 (the "License");
# you may not use this file except in compliance with the License.
# You may obtain a copy of the License at
#
#     http://www.apache.org/licenses/LICENSE-2.0
#
# Unless required by applicable law or agreed to in writing, software
# distributed under the License is distributed on an "AS IS" BASIS,
# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
# See the License for the specific language governing permissions and
# limitations under the License.
#
import webapp2
import json
import urllib
from google.appengine.api import urlfetch

class LoginInstagram(webapp2.RequestHandler):
 def get(self):
  self.redirect("https://api.instagram.com/oauth/authorize/?client_id=hoge&redirect_uri=http%3A%2F%2Ffoo.appspot.com%2Ftest%2Finsta%2Fredirect&response_type=code")
  

class RedirectFromInstagram(webapp2.RequestHandler):
 def get(self):
  self.response.headers['Content-Type'] = 'text/plain'
  code = self.request.get('code')

  url = "https://api.instagram.com/oauth/access_token"
  payload = urllib.urlencode({
   "client_id":"hoge",
   "client_secret":"hoge_secret",
   "grant_type":"authorization_code",
   "redirect_uri":"http://foo.appspot.com/test/insta/redirect",
   "code":str(code)
  })
  result = urlfetch.fetch(
   url=url,
   payload=payload,
   method=urlfetch.POST,
   headers={'Content-Type': 'application/x-www-form-urlencoded'}
  )
  if result.status_code == 200:
   #tokenの表示
   self.response.out.write(result.content)

app = webapp2.WSGIApplication([
 ('/test/insta', LoginInstagram),
 ('/test/insta/redirect', RedirectFromInstagram)
 ],debug=False
)

Oauth2だとこんなに簡単なものかとびっくりしてしまった。

ライブラリなどを使わずに実装できてしまうのである。

Oauth2は、どんどん普及してほしいなと思う昨今である。