Open Settings App from another App in iOS - React Native

Question

I'm going through the docs in React Native and can only find navigating to external links from the app I am in.

I want to be able to navigate to the Settings app (more specifically to the privacy > location services page) but, can not seem to find the necessary information on it. There is the native iOS way of doing it which I am trying to replicate through React Native.

Is this possible?

I have tried the following to detect if there is a Settings URL. The console logs that the Settings url works however, it does not navigate to that page.

Update: thanks to @zvona I am now navigating to the settings page but not sure how to get to a specific deep link.

Linking.canOpenURL('app-settings:').then(supported => {
            
    console.log(`Settings url works`)
    Linking.openURL('app-settings:'
}).catch(error => {
    console.log(`An error has occured: ${error}`)
})

Answer 1

You can access settings of the application with: Linking.openURL('app-settings:');

But I don't know (yet) how to open a specific deep-link.

2021 update use:

Linking.openSettings();

otherwise your app will be rejected due use of non-public URL scheme

Answer 2

I successfully opened the settings by the code below, hope it's helpful :)

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));

Reference: https://facebook.github.io/react-native//docs/linking.html

Answer 3

Since React Native version 0.59 this should be possible using openSettings();. This is described in the React Native Linking documentation. Although it did not work for me. When I tried quickly I saw a _reactNative.Linking.openSettings is not a function error message.

Linking.openSettings();