使用带有弹性豆茎的 spotipy oauth 的问题

如何解决使用带有弹性豆茎的 spotipy oauth 的问题

我在将 Flask 应用程序部署到 Elastic Beanstalk 时尝试使用 spipy 进行用户授权时遇到了这个问题。我收到错误消息，该页面在我的计算机上本地运行时运行正常，但响应时间过长：

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相比之下，这段代码确实有效（不同的是我没有尝试刷新这段代码中的令牌）：

from flask import Flask,request,redirect,render_template
from spotipy.oauth2 import SpotifyOAuth

import spotipy

app = Flask(__name__)

auth = SpotifyOAuth("xxxxxxxxx","xxxxxxxxx","http://mywebsite.com/callback",cache_path=".spotifycache",scope="user-library-read,playlist-modify-public")
sp = spotipy.Spotify(auth_manager=auth)
user = sp.current_user()

@app.route('/')
def index():
  return render_template('index.html')

@app.route('/my-link/')
def my_link():
  return render_template('result.html')

# Routes
@app.route('/my-link/',methods=['GET','POST'])
def playlist():

    token_info = auth.get_cached_token()
    if not token_info:
        # If there isn't a cached token then you will be redirected to a page where you will be asked to login to spotify
        # After that procceed to /callback
        auth_url = auth.get_authorize_url()
        return redirect(auth_url)

    token = token_info['access_token']

    # At this point you can now create a Spotifiy instance with
    spotipy.client.Spotify(auth=token)

    user = sp.current_user()
    return render_template('result.html')

@app.route("/callback")
def callback():
    url = request.url
    code = auth.parse_response_code(url)
    token = auth.get_access_token(code)
    # Once the get_access_token function is called,a cache will be created making it possible to go through the route "/" without having to login anymore
    return redirect("/")

if __name__ == '__main__':
    app.run()

提前致谢！