How to Open Amazon App from Within My App

How to open Amazon app from within my app?

So after some more research, based on https://www.appsight.io it seems that the amazon app does not use "amzn://" url scheme, but "amazonToAlipay://". After changing it to this, the UIApplication.shared opens the Amazon app.

Thanks to @LinusGeffarth and his answer to another related question.

How to open the Amazon AppStore directly from my Android application?

The Amazon app store URI is

amzn://apps/android?

Directing to a particular app would look like

amzn://apps/android?p=com.amazon.mp3

Source: Linking To the Amazon Appstore for Android

How to launch Amazon Shopping app in Android app?

I'd like to be able to launch the Amazon Shopping app from my android
application. How is this possible? What parameters would need to go
into the Intent?

You can use PackageManager#getLaunchIntentForPackage

startActivity(getPackageManager().getLaunchIntentForPackage("com.amazon.mShop.android.shopping"));

In addition, how would it be possible to pass a deep-link parameter so
that it lands on a specific product page?

It depends on whether Amazon app implements deep link and exposes intent-filter to external app. I assume it's not possible, but maybe you can ask Amazon.

How to launch Amazon app using product link in flutter?

Try below code hope its helpful to you.

You must use url_launcher package from here add this dependency in your pubspec.yaml file

You can refer my answer here also for open app from URL

create one widget

InkWell(
      hoverColor: Colors.transparent,
      child: Image.network(
        'https://upload.wikimedia.org/wikipedia/commons/d/de/Amazon_icon.png',
        width: 70,
        height: 70,
      ),
      onTap: () => amazon(),
    )

Create ontap function

amazon() async {
    const url =
        'https://www.amazon.com/';// or add your URL here
    if (await canLaunch(url)) {
      await launch(url);
    } else {
      throw 'Could not launch $url';
    }
  }

---

Related Topics