React-Native: How to open google play store from react native app?

Question

I am trying to find a way how to open the google play store with my application when the user is pressing a button inside of the react-native application. Is there a way how to do this?

Answer 1

If people are too lazy to read through the annoyingly SPA dev docs of android:

market://details?id=<package_name>

In react-native land:

Linking.openURL("market://details?id=<package_name>");

Example:

Linking.openURL("market://details?id=googoo.android.btgps");

Edit: Make sure to import Linking in order to use it

import { View, Text, StyleSheet, Linking } from 'react-native';

Answer 2

You can use deeplinking to redirect your user from your app using this: https://developer.android.com/distribute/tools/promote/linking.html

and the Linking API from react-native: http://facebook.github.io/react-native/docs/linking.html

Answer 3

For future users, you could go on to read the docs from the accepted answer for more info. But if you want a faster result, here it is.

Using Linking from react-native as

import { Linking } from 'react-native'

Linking.openURL("http://play.google.com/store/apps/details?id=<package_name>")

If you're using expo, then the below code is for you.

Install the expo-linking package via expo install expo-linking

import * as Linking from "expo-linking";

Linking.openURL("http://play.google.com/store/apps/details?id=<package_name>")

NOTE: <package_name> is your app's package name defined in the manifest file or app.json if you're using expo

Answer 4

Seems like this package is a cool alternative.

Available for React Native, and Expo

https://github.com/kimxogus/react-native-version-check

Answer 5

This worked for me: Go to google play. search your app there and copy the url and put like below

import { Linking } from "react-native";
Linking.openURL("your app url her")

Answer 6

Expo Linking docs mentions.

It's possible that the user doesn't have the Lyft app installed, in which case you may want to open the App / Play Store, or let them know that they need to install it first. We recommend using the library react-native-app-link for these cases.

It's got an MIT license so we're good to use it.

Let's see how that library does it!

https://github.com/FiberJW/react-native-app-link/blob/master/index.js

export const openInStore = async ({ appName, appStoreId, appStoreLocale = 'us', playStoreId }) => {
  if (Platform.OS === 'ios') {
    Linking.openURL(`https://apps.apple.com/${appStoreLocale}/app/${appName}/id${appStoreId}`);
  } else {
    Linking.openURL(
      `https://play.google.com/store/apps/details?id=${playStoreId}`
    );
  }
};

So I modified it to:

export const openInStore = async (config: {
    appStoreConfig: {
        appName: string;
        appId: string;
        storeLocale: string;
    };
    playStoreConfig: {
        appId: string;
    };
}) => {
    if (Platform.OS === "ios") {
        await Linking.openURL(
            `https://apps.apple.com/${config.appStoreConfig.storeLocale}/app/${config.appStoreConfig.appName}/id${config.appStoreConfig.appId}`
        );
    } else if (Platform.OS === "android") {
        await Linking.openURL(
            `https://play.google.com/store/apps/details?id=${config.playStoreConfig.appId}`
        );
    }
};

Usage:

try {
    openInStore({
        playStoreConfig: {appId: "com.stevemoretz.test"},
        appStoreConfig: {
            appId: "com.stevemoretz.test",
            appName: "Test",
            storeLocale: "us",
        },
    });
} catch (e) {}

Should work but haven't been tested yet!