XF 2.0 Adding new Connected Account Providers?

[Jaxel]

Looking at XF\ConnectedAccount\Provider and XF\ConnectedAccount\ProviderData, I've been able to create new files for my provider... but I'm not seeing where to actually change API endpoints... What am I missing here?

 

[Jaxel]

Running the "test provider" function, I am getting the following error:

InvalidArgumentException: Cannot find a valid OAuth Service for provider 'Discord' in src/XF/ConnectedAccount/Provider/AbstractProvider.php at line 152

1. XF\ConnectedAccount\Provider\AbstractProvider->getOAuth() in src/XF/ConnectedAccount/Provider/AbstractProvider.php at line 103
2. XF\ConnectedAccount\Provider\AbstractProvider->handleAuthorization() in src/XF/Admin/Controller/ConnectedAccount.php at line 68
3. XF\Admin\Controller\ConnectedAccount->actionPerformTest() in src/XF/Mvc/Dispatcher.php at line 249
4. XF\Mvc\Dispatcher->dispatchClass() in src/XF/Mvc/Dispatcher.php at line 89
5. XF\Mvc\Dispatcher->dispatchLoop() in src/XF/Mvc/Dispatcher.php at line 41
6. XF\Mvc\Dispatcher->run() in src/XF/App.php at line 1880
7. XF\App->run() in src/XF.php at line 328
8. XF::runApp() in admin.php at line 13

Where do I create an OAuth Service? I'm not finding any OAuth services for other providers to use an example.

 

[Z61]

Have you looked in src/vendor/lusitanian/oauth?

 

[Mike]

Note this comment in the abstract provider class for defining the service:

/**
 * Represents the name of the OAuth service.
 *
 * This can be given as a class name if there isn't a service already defined within the OAuth library.
 * e.g. '\XF\ConnectedAccount\Service\ProviderId' or 'XF:Service\ProviderId'
 * 
 * You can check if a service already exists by looking in 'src/vendor/lusitanian/oauth/src/OAuth/OAuth(1|2)\Service'
 *
 * @return string
 */

 

[Jaxel]

Oh... I get it now...

    public function getOAuthServiceName()
    {
        return 'EWR\Discord:Service\Discord';
    }
    
    public function getProviderDataClass()
    {
        return 'EWR\Discord:ProviderData\Discord';
    }