使用Flask和python-instagram库获取访问令牌时出现错误。任何想法将不胜感激
尝试获取授权网址时出现错误:
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Web开发的新手,我认为我可以通过创建一个简单的应用程序来学习,该应用程序可以根据标签等不同的搜索条件提取IG图片。任何帮助都会有用。
答案 0 :(得分:0)
TypeError: The view function did not return a valid response. The function either returned None or ended without a return statement.
该错误实际上告诉了您所有需要了解的内容。您需要在make_response()
中返回一些内容(假设url变量正确无误,并且您有一条路由)。
try:
url = api.get_authorize_url(scope=None)
return redirect(url_for(url))
except InstagramAPIError as e:
print(e)
您可以阅读有关此here的更多信息。