Reputation: 12084
I'm creating an react-native app and I want to show and warning to user if their IOS version is < 11.0.
That works fine. I have something as follows:
What I want is to go the phone settings if they click on Go to settings.
The code for that is as follows:
Linking.canOpenURL('app-settings:').then(supported => {
if (!supported) {
console.log('Can\'t handle settings url');
} else {
return Linking.openURL('app-settings:');
}
}).catch(err => console.error('An error occurred', err));
Thus I try to do it with Linking.openURL('app-settings:')
. But the problem is that that goes to the settings of my app, but I want to go to the settings
index page or even better to system upgrade
setting.
I need it only for IOS.
Any idea how can I do that?
Upvotes: 3
Views: 2489
Reputation: 6992
You can no longer do that. You could use the URL App-Prefs:
, which does what you want, but that will cause your app to be rejected upon review, because it's considered non-public URL scheme.
The only URL allowed by Apple is the constant UIApplicationOpenSettingsURLString
, which takes you to your app's setting, not general settings of the device.
Upvotes: 5