I want to share URLs from the browser to my React Native app and open them in the app, how can I do that?
This is called deep-linking, that your app launches and even does some actions depending on the called URL of your app. Many apps use it that's how you probably know it.
You can read about this and implement it.
Here is some articles and documentation about it:
https://reactnavigation.org/docs/deep-linking/
https://medium.com/react-native-training/deep-linking-your-react-native-app-d87c39a1ad5e
Related
I am trying to implement deep linking into this app. The app is a React Native Expo app, and is served on web, iOS and Android. There are a bunch of things we want to achieve with this i.e. link to specific things in the app such as certain messages etc. But currently my primary focus is just to navigate to general screens like 'Home', 'Messages', 'Login' etc.
I have been testing and so far, I can enter the URIs such as exp://host:port/--/login and they work as expected.
However, I know and understand that this will not work in a web browser on my laptop for example.
So my question is, how do I send an email to a user let's say after they have verified their account, prompting them to login, with a link that will for one, take them to the web app if they are on desktop, and two open the mobile app if they have it installed on their phone?
And just as an example, let's imagine my deep link URI is myapp://login
Thanks for any help.
EDIT
After looking around some more, I am deciding to consider the possibility of linking to the web app, and upon the web app loading, linking to the mobile app if they have it installed.
Given this, what is the best way to handle this?
You can follow one of the next approaches.
Validate the request origin and based on that generate the link for
an specific platform.
Generate as much links as platform are currently serving you app.
Like: Link to iOS, Link to Android and Link To web
You can design image buttons for each one to make it prettier
I'm trying to communicate two React Native apps using Firebase Dynamic Links, only in Android.
When I execute openLink with the URL of the other, I see the browser for a second, and then it opens the other app well.
I don't want to see an intermediate browser before open the other app.
I'm having that issue from app A to B, and viceversa.
So, why is the browser opening first? And how can I configure the apps in order to not open the browser?
it's default behavior of android devices.
usually if any link supported by app and its set default to open link then it will open directly otherwise android system opens browser and based on Link URI scheme navigate to app.
there is one solution,
You can create module (intent activity) which will launch that app directly.
Steps to follow
pass data from js to native module and from that use Intent class, set data and start it.
this was for Android use case.
I am wondering if it is possible to integrate TWA (Trusted Web Activities) into an existing React Native project. This way I could have a section in my app where costumers can use my PWA inside my app. As fallback for iOS I would use something like their WebView. If it is possible, how would I go about implementing it?
In short, yes, it is possible to integrate Trusted Web Activity into a React Native App.
You'd need to create an Android Native Module that wraps Android Browser Helper into a React Native API. Since you want to use the Trusted Web Activity as part of your app, you will probably be looking into wrapping TwaLauncher.
Then, you will need to implement a module for iOS, which will have the same API in React, but will use the WebView as an implementation.
It seems someone has already created a wrapper for Android (but I haven't tested).
I am trying to open an external application from a react native application. I passed the URL and it calls the application but it shows loading screen only and doesn't show the application.
Could anyone suggest the possible reason?
I tried using react-native-app-link which seems to be outdated and not useful. Then I tried linking.openurl(WhatsApp://) but it didn't work as well
Linking.openURL("whatsapp://app")
The external app should open.
if you would like to open only Whatsapp through your application then
refer https://faq.whatsapp.com/en/iphone/23559013
thank,
hope this helps you
Using onesignal, When one sets the push notification value "url" for a notificaiton to open in an app, aka universal linking, by default the app is opened and a webview screen emerges (not on android, only iOS - at least in React Native).
I want to ignore this webview. I don't want to open it. The url is supposed to open the app to the corresponding universal link/deep link, not my domain on a web broswer. If I click on a similar link from a non-notificaiton source, it takes me to the screen, why does the push notification not do this?
How do I override this? I can't seem to figure it out. Do I have to manually implement this? Why does it open a webview by default and not the corresponding screen in the app instead?
It seems you have to manually route it if one were to go this route, unfortunately. From what I'm gathering bits here and there.
this is a bad choice to go with if you're using iOS:
https://www.myapp.com/p/fdsafa
this is better:
myapp://p/fdsafa
unfortunately onesignal has it setup in some weird way with the https://www.myapp.com that it makes it useless on iOS, unless your actual intention is to open an actual web url in a webview. Where as myapp:// is compatible with both platforms (just don't put a host on Android, leave the host out, just do the myapp as the schema). Works.