Opening Safari settings with NSURL Swift

Question

Is it possible to open Safari app settings using NSURL link ?

Here is the code i'm using for general settings

UIApplication.sharedApplication().openURL(NSURL(string: UIApplicationOpenSettingsURLString)!)

How should i modify it - if its even possible?

Thank you!

score 2 · Accepted Answer · edited May 23 '17 at 12:14

2

From that post follow this two steps:

First go to Info -> URL Types and add URL Schemes as prefs as shown in below image:

After that use this code for open safari settings:

UIApplication.sharedApplication().openURL(NSURL(string: "prefs:root=Safari")!)

And result will be:

edited May 23 '17 at 12:14

Community

answered Sep 25 '15 at 10:34

Dharmesh Kheni

sorry, Dharmesh, but this approach leads to the AppStore rejection ( – David Robertson Sep 30 '15 at 22:39

kakaiikaka · Answer 2 · 2023-05-13T12:58:53.207

As of iOS 16, should be

Link("Setting Safari", destination: URL(string: "App-Prefs:Safari")!)

And we can even goto the subpages with an &path=WEB_EXTENSIONS

Link("Setting Safari", destination: URL(string: "App-Prefs:Safari&path=WEB_EXTENSIONS")!)

Also if you want goto somewhere else in the settings, it is also possible:

taking Apple ID page as an example:

Link("iCloud Account", destination: URL(string: "App-prefs:APPLE_ACCOUNT")!)

BTW, on our app, the review passed.

see github repo for more examples.

2 Answers2