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issue with authorizing credentials #718
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Try opening your sheet by key using |
Hi @CubeBlazer, thanks for the response. The issue doesn't seem to be with opening the sheet. It seems to be in authorizing using my credentials ( i.e. gc = gspread.authorize(credentials) is the failure point based on the error.) |
FWIW having the same problem, also on PythonAnywhere. It runs on my local machine though... |
I am also having auth issues. `#!/usr/bin/env python3 import gspread After following the Read the Docs page here
I noticed in the documentation that it refers to a "Service Key" as opposed to a Service Account- is it possible the documentation is out of date and there are further steps to the auth process? Edit: I have left other endpoints I tried to auth against in the code. This is running from my local machine, Python 3.7 on Debian 10. pip3 list certifi 2019.11.28 |
My issue was fixed by running |
check this link. it resolve my same problem |
Hey all, I was having similar OAuth issues when trying to update some old code (using gspread==0.6.2) to use the latest versions (gspread>=3.0.0) and I believe it might be the same solution required here. As per the Google Sheets API docs: https://developers.google.com/sheets/api/reference/rest/v4/spreadsheets/get#authorization-scopes
Hope that helps someone else too. TL;DR: Change |
This worked for me as well! Had the same issue as OP while running on PythonAnywhere. |
Issues seems resolved. we now use the new scopes for the API V4 and it's all provided by default by gspread suing closing this issue. |
I am running the basic authentication process (after generating the OAuth credentials outlined in the documentation: https://gspread.readthedocs.io/en/latest/oauth2.html
Here is my code for reference:
I am receiving a type error, saying the argument for the authorize function in my code is a string. Has anyone seen this issue? I was not able to find any other issues where this was replicated. I am using python 3.7 and am running this code on python anywhere: https://www.pythonanywhere.com/ Any ideas would be greatly appreciated.
Full Error:
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