Issue
I am developing a mobile application using React Native in which I want to open mobile device’s settings when user clicks on a specific button.
I have gone through the docs of RN, but still couldn’t find anything that can help me in achieving the same. Can someone guide me on the same, how can that be implemented?
Thanks 🙂
Solution
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();
}}
/>
);
Answered By – CR7
Answer Checked By – Mary Flores (FlutterFixes Volunteer)