I want to know how to open an external application to the current one by pressing a button, because I was commissioned to create an application that makes changes in a database and then another already developed one is executed. I don't know if it's possible but I hope you can help me. From already thank you very much.
I tried with expo-linking, react-native-app-link, OpenAplication and rn-openapp. With OpenAplication it gives me an error that causes the application to close and in the others I didn't get good results, maybe it's that I didn't understand the documentation well or I don't know.
There is the code that I use and below I attach the app that I want to open, if you need anything else tell me.
import React from "react";
import { StatusBar } from "expo-status-bar";
import { StyleSheet, Text, TouchableOpacity, View } from "react-native";
import { MotiView } from "moti";
import * as Linking from "expo-linking";
const App = () => {
return (
<View>
<StatusBar hidden={false} />
<View style={{ marginTop: 150 }}>
<Text style={{ fontSize: 25 }}>Cargando...</Text>
<TouchableOpacity
onPress={() => {
Linking.openURL("https://stackoverflow.com/")
.then((res) => {
console.log(res);
})
.catch((err) => {
console.log(err);
});
}}
>
<Text>Boton</Text>
</TouchableOpacity>
</View>
</View>
);
};
export default App;
const styles = StyleSheet.create({
shape: {
justifyContent: "center",
height: 100,
width: 100,
borderRadius: 25,
marginRight: 10,
backgroundColor: "black",
},
container: {
flex: 1,
alignItems: "center",
justifyContent: "center",
flexDirection: "column",
backgroundColor: "#fff",
},
});
