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I want to open my Android app (specific activity) when user clicks on a link or a website redirect to that link. ex link: com.abc.efg/callback?wjiofjjfie... I have used this primitive way by adding intent filter to that particular activity but it does not work. everything opens in browser.

<activity
            android:name=".UrlActivity"
            android:screenOrientation="portrait"
            android:exported="true">
            <intent-filter>
                <action android:name="android.intent.action.VIEW" />
                <category android:name="android.intent.category.DEFAULT" />
                <category android:name="android.intent.category.BROWSABLE" />

                <data android:scheme="https"/>
                <data android:host="com.abc.efg"/>
                <data android:scheme="http"/>
                <data android:host="com.abc.efg"/>

            </intent-filter>

        </activity>

i have also searched a lot and found on android documentation that from android 12 and later **we have to make this url an App Link ** refer- text so, is their another way to open the app when link clicked without making the link App link? If not, than how can I make this link an App link?

Abhi
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  • From Android 12 your app needs to verify domain. There is an [answer](https://stackoverflow.com/questions/71079961/how-to-verify-android-app-links-on-android-12-and-higher) – Injent Jul 15 '23 at 03:55
  • actually, the problem is the redirection is coming from authentication server and also the redirect has token id and access token that i need to pass in my app and in dynamic link it is not possible to add the parameter in the redirect link in the firebase dynamic link format – Abhi Jul 15 '23 at 12:46

1 Answers1

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From high version Android you can use Firebase dynamic link , but this will be deprecated soon maybe next 12 months, because I heard Google noticed that

Nguyễn Minh Khoa
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