Yevhen Dashkevych

Fix KeyError: 'graphql'

1

Traceback (most recent call last): File "C:\***\site-packages\Instagram\agents.py", line 67, in update data = data[key] KeyError: 'graphql' ................... raise UnexpectedResponse(exception, response.url) Instagram.exceptions.UnexpectedResponse: Got unexpected response from 'https://www.instagram.com/p/xxxxxxxxxxx/' Error: 'graphql'