android - open android application using url and go to different screen

Question

how to start my app using a url link and go to different activity. i have used this code

<activity
            android:name=".Act1"
            android:label="activity first"
            android:screenOrientation="landscape" >
            <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:host="xyz.com"
                    android:pathPrefix="/abc/"
                    android:scheme="https" />
            </intent-filter>
        </activity>

by using this code i am able to open Act1 while entering into my android app but i want to switch to Act2 and Act3. because i have to use three different url and for all the stating screen should be Act1, Act2 and Act3 as click on url1, url2 and url3 respectively. please suggest if any solution.

Answer 1

最好的办法是启动相同的活动（“前台活动”）。在这个“前端活动”中，您解析 url 并根据 url 启动 Act1、2、3。