Groffer Anderson

Thanks Trapta Singh & Ifra Saqlain,
These solutions worked like a charm. 

Hi, 
I'm looking for a solution where I'll be showing an option on the Zendesk Guide (Help Center) Template on the header only if the user is an Admin or Agent. However, I'm unable to find the right variable in the template code that shows me the user type. 

In the header.hbs code, we have an option to check if the user is signed in or not by variable "signed_in", but we have nothing to check the user type. 

One detail I found on the "user_profile_page.hbs" is that it shows user.admin and user.agent, these variables are not accessible on the header. Is there a solution to pull the user Type? 

Ref Image of code checking if the user is signed in:

Thanks @...
So, The API does support deleting a session if I have a session_Id. Is there anyway to get the session_id when I login on the Zendesk console using the OAuth Grant? Or do we have a case where I can fetch the current session of zendesk, If I've used OAuth Grant flow? 
If i won't have the session Id, I'll have to delete all the running sessions of the user, which does not seem to be the best option. 

We have a custom portal where our users login. The entire login process is the Authorization grant flow and once Logged in, the browser has the session stored for Zendesk and I have the Auth Token to make API calls. However, once the user logs out from the custom portal, we revoke the Auth token, but the browser sessions are still valid and they can still access their logged in Zendesk. 

Question:

1. How do I logout the local session also of the user when I click on the Logout option on the custom portal?

2. I've also tried pulling the Active sessions from the API, but It gives me all the active sessions. It becomes really difficult to find the current session that has to be deleted from all the list of active sessions. I don't want to logout the user of a wrong session.

Additional Details: 

Currently I'm using the Logout URL ( company.zendesk.com/access/logout ) to Logout user. I open this link in a new tab to clear the session.  But I don't think this is the best possible way. 

Is there a proper way to logout users? Deleting the sessions looks the most promising way but when I have multiple sessions active, and I have to only delete a single (the current session) one, It becomes difficult to decide which one is the current local session. 

References: 
1. Zendesk Sessions API
2. Zendesk Logout URL

Hey @...
Thanks for the update. 

I'm working with the Authorization Grant flow from Zendesk to Authenticate users into my Custom Website. Once I authenticate them, a page to grant permissions appears, where users can Allow/Deny my webapp to access details from their Zendesk. 

Questions :

1. I wish to change the Content being displayed here. I really wanted to know if I can rephrase the text details like "Read all user data" to something more subtle for my customers. 
2. In the tutorial for zendesk Auth Grant flow, The same Auth Grant page looks really different that is being used by an application ("Asana"). Is there a way I can similarly customize my Auth Grant page like this? 

Detailed Imaged:

My Authorization page:

Asana's Auth page:

References: 

1.  Zendesk Authorization grant flow.
2.  Zendesk Authorization grant flow tutorial

Hi @...
Can you help me out with a similar problem here : https://support.zendesk.com/hc/en-us/community/posts/1260801978190-How-to-perform-Authorization-code-grant-flow-on-Zendesk-with-Agent-when-I-have-SSO-enabled-

Thanks for the update @.... 

How can I extract OAuth tokens for my users (zendesk agents) if they're all connected via SSO? I'm using Okta (SAML) to SSO into Zendesk. It's not possible for me to get the API tokens for all my agents. What is the best way to get the OAuth tokens for these agents? 

Note: I have a custom login, where my agents login using their OKTA credentials. I can hold these credentials and use them to get the Zendesk token as well, if there is a way to get my Auth token for the agent. (using their credentials) .