使用 python 获取 Instagram 访问令牌

Question

我在使用 Flask 和 python-instagram 库获取访问令牌时出错。任何想法将不胜感激

尝试获取授权时出现错误 url:

Traceback (most recent call last):
  File "C:\Python37\lib\site-packages\flask\app.py", line 2309, in __call__
    return self.wsgi_app(environ, start_response)
  File "C:\Python37\lib\site-packages\flask\app.py", line 2295, in wsgi_app
    response = self.handle_exception(e)
  File "C:\Python37\lib\site-packages\flask\app.py", line 1741, in handle_exception
    reraise(exc_type, exc_value, tb)
  File "C:\Python37\lib\site-packages\flask\_compat.py", line 35, in reraise
    raise value
  File "C:\Python37\lib\site-packages\flask\app.py", line 2292, in wsgi_app
    response = self.full_dispatch_request()
  File "C:\Python37\lib\site-packages\flask\app.py", line 1816, in full_dispatch_request
    return self.finalize_request(rv)
  File "C:\Python37\lib\site-packages\flask\app.py", line 1831, in finalize_request
    response = self.make_response(rv)
  File "C:\Python37\lib\site-packages\flask\app.py", line 1957, in make_response
    'The view function did not return a valid response. The'
TypeError: The view function did not return a valid response. The function either returned None or ended without a return statement.

这是失败的代码：

  try:
    url = api.get_authorize_url(scope=None)
    redirect(url)
  except InstagramAPIError as e:
    print(e)

背景： 我是 python 和网络开发的新手，我想我会通过创建一个简单的应用程序来学习，以根据不同的搜索条件提取 IG 图片，例如标签。任何帮助都会有用。

Answer 1

TypeError: The view function did not return a valid response. The function either returned None or ended without a return statement.

这个错误确实告诉了你需要知道的一切。您需要 return make_response() 中的某些内容（假设 url 变量是正确的并且您有它的路线）。

  try:
    url = api.get_authorize_url(scope=None)
    return redirect(url_for(url))
  except InstagramAPIError as e:
    print(e)

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