Hi,
I have developped a Watch App (with an accompanying iOS app that does nothing but host the watch app - no functionality of its own) which is a fully featured Trello client.
To login to the watch app, I show a random six character code and direct the user to login using a computer, tablet or phone to https://login.myapp.com/ (not the real URL) , where they are logged in using Trello's authentication and then enter the code shown on the watch. They watch app then logs them in automatically without doing anything futher on the watch.
I got rejected twice with:
Guideline 4.0 - Design
We noticed that the user is taken to Safari to sign in or register for an account, which provides a poor user
experience.
Next Steps
To resolve this issue, please revise your app to enable users to sign in or register for an account in the app.
We recommend implementing the Safari View Controller API to display web content within your app. The Safari View
Controller allows the display of a URL and inspection of the certificate from an embedded browser in an app so that
customers can verify the webpage URL and SSL certificate to confirm they are entering their sign in credentials
into a legitimate page.
Resources
For additional information on the Safari View Controller API, please review the What's New in Safari webpage.Watch apps can't use Safari and I can't implement a Safari View Controller in a Watch App...
I got rejected for the same reason earlier, and submitted an update with these comments:
I've just submitted an update (build 16) which clarifies the login message to be clear that the app has no native
login/account creation capability. Authentication occurs using Trello, a third-party system. When prompted please
login using a web browser on a phone, tablet or computer. There is no way for me to allow login directly on the
watch - this is technically impossible.
Your previous rejection included suggestions which are not available for WatchOS apps
("We recommend implementing the Safari View Controller API to display web content within your app").I got rejected for the same reason with no acknowledgement as to my previous response. Any suggestions?
Damian
