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String is not JSON formatted
Tags: app administration
Jun 3, 2021 at 9:43am   •   3 replies  •  0 likes
Rick Smith

getting exception string is not JSON formatted


This is throwing it, what should this be?

I have been using

RINGCENTRAL_SERVER = 'https://platform.devtest.ringcentral.com'

rcsdk = SDK(os.environ[RINGCENTRAL_CLIENTID], os.environ[RINGCENTRAL_CLIENTSECRET], os.environ[RINGCENTRAL_SERVER])

And it worked, has this changed?


I have also tried
RINGCENTRAL_SERVER = 'https://platform.devtest.ringcentral.com/restapi/oauth/token'


No difference


Using Python 3.7 64




on Jun 3, 2021 at 9:25pm   •  0 likes

Please revise your question, I really don't know what worked and what did not work.

on Jun 4, 2021 at 5:54am   •  0 likes

Hello,


Thanks for the response, This is what I am getting now, it seems to object to the client id but it does not say why , I have copied this ID directly from the account page...waht am I doing wrong?

Thanks

Message='1MQ8Fg52RwOZCcJWrA4-kA'
Source=C:\Users\Rick\source\repos\RCMaintNotify\RCMaintNotify\RCMaintNotify.py
StackTrace:
File "C:\Users\Rick\source\repos\RCMaintNotify\RCMaintNotify\RCMaintNotify.py", line 41, in <module>
rcsdk = SDK(os.environ[RINGCENTRAL_CLIENTID], os.environ[RINGCENTRAL_CLIENTSECRET], os.environ[RINGCENTRAL_SERVER])



2 Answers
answered on Jun 5, 2021 at 7:14am  

So tried changing the password?? No go!, It just does not work, I need to think some about going forward with this...I am still not sure what the exception is? Other than it still objects to the client ID...


 0
answered on Jun 4, 2021 at 7:21am  

Why don't you try with the value first to see if all the parameters are correct

RINGCENTRAL_CLIENTID = '1MQ8Fg52RwOZCcJWrA4-kA'
RINGCENTRAL_CLIENTSECRET = 'abc...'
RINGCENTRAL_SERVER = 'https://platform.devtest.ringcentral.com'

rcsdk = SDK(RINGCENTRAL_CLIENTID,RINGCENTRAL_CLIENTSECRET,RINGCENTRAL_SERVER)

If that works, then the problem is in your environment. If that does not work, then those params could be incorrect, but let's see what you'll get first.


 0
answered on Jun 4, 2021 at 8:01am  

No change it is still objecting to: '1MQ8Fg52RwOZCcJWrA4-kA'


 0
on Jun 4, 2021 at 8:25am   •  0 likes

Looking at your app settings, it is a "Public" app with "Desktop/Windows" platform type. This type of app does not support password flow authentication. Is there a reason you choose those settings for your app? To use password flow, you must choose the app platform type as "Server only- No UI" and it must be a "Private" app.

You cannot modify this app anymore. Either you create a new app properly or let me know if you want to change this app. I can help to change those settings.

on Jun 4, 2021 at 8:55am   •  0 likes

Not sure why that was done, I am inheriting this...If you can help change it, I would appreciate it. Otherwise I will just create a new one.


Thanks|
Again

on Jun 4, 2021 at 8:57am   •  0 likes

It's now Private and Password Flow type. Try again

on Jun 5, 2021 at 6:59am   •  0 likes

Nope, no joy same exception, I have tried creating another new app, with the settings you specified, no change, I noticed when I tried creating a new app with those settings password is not available??


maybe I should change the password? Running out of ideas...

on Jun 5, 2021 at 7:09am   •  0 likes

Check if you can login https://service.devtest.ringcentral.com with that user credentials. Then you must use the same user credentials to login your app.

on Jun 5, 2021 at 7:38am   •  0 likes

Yes, I was able to get in with those credentials

on Jun 7, 2021 at 7:13am   •  0 likes

You new app "ChiliSMS" still has wrong platform type and access type. I recommend you read and use the dev guide getting started help. Create a new app by clicking the "Create SMS app" button.

https://developers.ringcentral.com/guide/messaging/quick-start



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