How can I open a specific page of another app from my app in flutter? - flutter

I’m trying to build a functionality where i want to open the settings page of WhatsApp from a simple onTap gesture in my app. To launch WhatsApp I’m using this dependency:
https://pub.dev/packages/device_apps
However it isn’t helping me open the settings page specifically.
Is there a tool available for me to do this?

You can use this by passing the package name of the another application.
dependencies:
android_intent: ^0.3.5

WhatsApp has a URL scheme to launch the app and send a message to a phone number.
iOS Url Scheme
Android Url Scheme
But I don't think that you will be able to launch WhatsApp settings directly.

Related

How to suggest Flutter Mobile app on clicking flutter web app url?

How to succeed When user tap on Flutter web url we should suggest whether open in Chrome or Flutter installed app?
If you want to open a native application selection dialog, then you should enable deep linking for your application.
This will result with an app chooser dialog on Android when user opens a URL which mobile application has registered as deep link or a "Open in application X" banner in Safari for iOS
In Flutter , I have used uni_links package and followed document to succeed

Flutter: how to open an user's Airbnb app inbox in guest mode?

I want to open my app's user's Airbnb inbox in guest mode.
Presently I'm launching the URL:
https://www.airbnb.co.uk/guest/inbox/
But it opens their Inbox in a browser, even if the App is present. Dissappointing the app doesn't redirect it, like Googlemaps does.
Is there a way to trigger the Airbnb app? A deeplink? How would I test if the app was present before launching it?
One of Airbnb's open source contributions was 'DeepLinkDispatch', so internally they're using Deeplinks a lot.
For that link and my other links I'm using the Flutter plugin url_launcher. Would be great to have a deep link URL to throw at that, if it fails I could launch the URL above as a fallback.

Send my app programmatically in flutter app

I have this flutter app fully done but I want to add a button that say share when user clicks on this button it share the entire apk to a friend either using Xender or any other sharing medium.
I kno it may not be possible on iOS but I want to do this on flutter Android.
How can this be achieved.
Why don't you upload to apk to any storage hosting then send the link when click share.
Another way is to use Flutter share package , you can take a look at the documentation for more info

Instagram in-app browser does not support add to home screen feature for pwa

https://caniuse.com/web-app-manifest
https://developer.mozilla.org/en-US/docs/Web/API/BeforeInstallPromptEvent
Here it says, android webview supports add to home screen. Even then in instagram's in-app browser it doesn't work.
Even the custom install that we trigger using beforeinstallprompt also doesn't show up.
Not any of the documentations mention anything about this.
In IOS there's a way you can install web apps using profile configuration files. Do you know if there is a way to do the same in Android? If so, we can redirect the bio link to an install page of our web apps.

how to open another app(my own app) from Flutter app?

I have two applications. customer app and vendor app. How to open vendor app from customer app when click button?
Is it possible to open another application from the Flutter app?
com.demo.customer and com.demo.vendor
I tried url_launcher. but It cannot give package name and open another application.
And also I tried this package,device_apps but not working for ios.
Should be work on Android and ios.
You need to register a custom url scheme for the app you want to open.
You can read more here:
For iOS: iOS Documentation
For Android: Android Documentation
There's also a nice plugin that can handle deeplink actions for Flutter.
Flutter Uni Links
When you have registered a custom url scheme (for example: demo://vendor) you can just call that url with url_launcher plugin:
launchUrl('demo://vendor')