How to Open Phone Settings When a Button Is Clicked

How do I open phone settings when a button is clicked?

Using UIApplication.openSettingsURLString

Update for Swift 5.1

 override func viewDidAppear(_ animated: Bool) {
    let alertController = UIAlertController (title: "Title", message: "Go to Settings?", preferredStyle: .alert)

    let settingsAction = UIAlertAction(title: "Settings", style: .default) { (_) -> Void in

        guard let settingsUrl = URL(string: UIApplication.openSettingsURLString) else {
            return
        }

        if UIApplication.shared.canOpenURL(settingsUrl) {
            UIApplication.shared.open(settingsUrl, completionHandler: { (success) in
                print("Settings opened: \(success)") // Prints true
            })
        }
    }
    alertController.addAction(settingsAction)
    let cancelAction = UIAlertAction(title: "Cancel", style: .default, handler: nil)
    alertController.addAction(cancelAction)

    present(alertController, animated: true, completion: nil)
}

Swift 4.2

override func viewDidAppear(_ animated: Bool) {
    let alertController = UIAlertController (title: "Title", message: "Go to Settings?", preferredStyle: .alert)

    let settingsAction = UIAlertAction(title: "Settings", style: .default) { (_) -> Void in

        guard let settingsUrl = URL(string: UIApplicationOpenSettingsURLString) else {
            return
        }

        if UIApplication.shared.canOpenURL(settingsUrl) {
            UIApplication.shared.open(settingsUrl, completionHandler: { (success) in
                print("Settings opened: \(success)") // Prints true
            })
        }
    }
    alertController.addAction(settingsAction)
    let cancelAction = UIAlertAction(title: "Cancel", style: .default, handler: nil)
    alertController.addAction(cancelAction)

    present(alertController, animated: true, completion: nil)
}

How to open Settings of Android Phone on a button click in our Android App

this should do it

startActivityForResult(new Intent(android.provider.Settings.ACTION_SETTINGS), 0);

starts settings activity for result

Going to Phone Settings when a button is clicked

The solution is to add

startActivityForResult(new Intent(Settings.ACTION_LOCALE_SETTINGS),0);

to your button's function or OnClick event.

Open phone settings when button is clicked in my app

Swift 3 iOS 10

let settingsUrl = NSURL(string:UIApplicationOpenSettingsURLString) as! URL
UIApplication.shared.open(settingsUrl, options: [:], completionHandler: nil)

SwiftUI: How do I open the (App-)phone settings when I push a Button?

Based on this, you can use the following sample code:

import SwiftUI

struct ContentView: View {
    var body: some View {
        Button(action: {self.settingsOpener()} ){
            Text("Open Settings")
            
        }
    }
    
    private func settingsOpener(){
        if let url = URL(string: UIApplication.openSettingsURLString) {
            if UIApplication.shared.canOpenURL(url) {
                UIApplication.shared.open(url, options: [:], completionHandler: nil)
            }
        }
    }
}

Android Studio: I want my button to open the phone settings

You need to start a new Intent. Here is an example to open the Wifi settings:

startActivity(new Intent(android.provider.Settings.ACTION_WIRELESS_SETTINGS));

You get go through the android.provider.Settings constants for different settings that you can go to.

Hope this helps.

How to open your app in Settings iOS 11

Here is the code you're looking for, I guess:

if let url = URL(string: UIApplicationOpenSettingsURLString) {
    if UIApplication.shared.canOpenURL(url) {
        UIApplication.shared.open(url, options: [:], completionHandler: nil)
    }
}

And in addition, the updated version for swift 5 :

if let url = URL(string: UIApplication.openSettingsURLString) {
    if UIApplication.shared.canOpenURL(url) {
        UIApplication.shared.open(url, options: [:], completionHandler: nil)
    }
}

React Native: How to open OS settings in Android and iOS?

Did you try to use Linking component form react-native (v0.60+)?

import { Button, Linking } from "react-native";

const Screen = () => (
  <Button
    title="Open Settings"
    onPress={() => {
      Linking.openSettings();
    }}
  />
);

How to open Settings programmatically like in Facebook app?

You can't, there is no API call to do this.

Only system dialogs, dialogs from Apple Frameworks, can open the settings app.
In iOS 5 there was a app url scheme to open the system dialog but Apple removed it later.

With the coming of iOS 8 you can open the settings dialog on your apps page.

if (&UIApplicationOpenSettingsURLString != NULL) {
   NSURL *url = [NSURL URLWithString:UIApplicationOpenSettingsURLString];
    [[UIApplication sharedApplication] openURL:url];
}
else {
  // Present some dialog telling the user to open the settings app.
}