How to open device setting for iOS8?

Question

I keep seeing that you can use

if UIApplication.shared.canOpenURL(settingsUrl) {
    UIApplication.shared.open(settingsUrl, completionHandler: { (success) in
        print("Settings opened: \(success)")
    })
}

but it only works for iOS 10 and above. But how can I do the same thing for iOS 8?

May be duplicate of https://stackoverflow.com/questions/5655674/opening-the-settings-app-from-another-app — Sateesh Yemireddi, Oct 18 '18 at 10:02

Satish · Answer 1 · 2018-10-18T10:17:20.710

0

UIApplication.openSettingsURLString will return settings url string.

if let settingsURL = URL(string: UIApplication.openSettingsURLString),
    UIApplication.shared.canOpenURL(settingsURL) {
    UIApplication.shared.open(settingsURL, completionHandler: { (success) in
        print("Settings opened: \(success)")
    })
}

edited Oct 18 '18 at 10:17

answered Oct 18 '18 at 09:59

Satish

2,015
1
14
22

open(_:options:completionHandler:) is only available on iOS 10.0 or newer – rminaj Oct 19 '18 at 01:04

score 0 · Answer 2 · answered Oct 18 '18 at 10:08

At the moment, I rarely see any application support for iOS 8 anymore. But if you want to know, here is the answer:

let urlStr = "yourURL"
if let linkURL = URL(string: urlString), UIApplication.shared.openURL(URL) {
 UIApplication.shared.openURL(linkURL)
}

score 0 · Answer 3 · answered Oct 19 '18 at 01:11

0

This is what worked for me

if UIApplication.openSettingsURLString != nil{
    if let aString = URL(string: UIApplication.openSettingsURLString){
        UIApplication.shared.openURL(aString)
    }
}

answered Oct 19 '18 at 01:11

rminaj

560
6
28

How to open device setting for iOS8?

3 Answers3