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I have a working Django AllAuth install with Google signin setup.

I'm now trying to set up Google Assistant integration for my app, using the Actions on Google console.

enter image description here

Documentation on OAuth2 account linking at https://developers.google.com/actions/identity/account-linking).

Here's a screenshot of my setup so far: actions on google set up screen

I'm using as authorization url:

djangodomainwithallauth.com/accounts/google/login/

I'm using as token url:

djangodomainwithallauth.com/accounts/google/login/callback

I've also set

LOGIN_REDIRECT_URL = 'https://www.google.com/?result_code=SUCCESS'

as per the actions on google doc, but it seems you can pass it with a next parameter.

I'm at the point where I can't login with the Google account chooser (fails silently), but I can login sucessfully to my account within the webview (but redirecting to https://www.google.com/?result_code=SUCCESS doesn't do a thing)

I'm not a seasoned programmer, what am I missing ?

PierreS
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1 Answers1

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Just to be clear, are you using Google Sign In credentials for the Add Quick Account Linking part? If not, you should create one following the steps here and use it for that part.

When you say it fails silently, what do you mean? Does anything pop up at all when you are using Quick Account Linking?

It also may be helpful to know that currently, you might have to disable and reenable device testing to switch between the redirect webview and Quick Account Linking when testing on device.

Shuyang Chen
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  • * Already using Google Sign In. * The Account picker pops in, but selecting a Google account fails (the popup just disappears and no request done on the server) * There's the option of loging in from a webview, but although I'm log successfully, Google does not recognize it as a connexion success. – PierreS Aug 24 '17 at 18:20
  • So, when you reinvoke the app, does nothing different happen? The current expected flow for *Sign in required for welcome intent* is to first invoke the app to perform the sign in, then after you are signed in, the second time you invoke it, it should not require sign in and send the accessToken to the app if successful. – Shuyang Chen Aug 24 '17 at 22:56