How to open my application when app link is clicked?

问题

What I want is to share the link of specific page of my application. Suppose I am in a profile page of some user in my app and I want to share that user to some of my friends so I would like to share it via WhatsApp and other applications.

So basically I want to share a link to my app.

This is what I am doing right now

Intent shareIntent = new Intent(android.content.Intent.ACTION_SEND);
    shareIntent.setType("text/plain");
    shareIntent.putExtra(android.content.Intent.EXTRA_TEXT, "https://play.google.com/store/apps/details?id=" + activity.getPackageName());
    activity.startActivity(Intent.createChooser(shareIntent, "Share"));

I know this is not what I really want but it's partially done. if user hasn't installed my app then this code redirect him to google play store page to download this app.

Main problem is when user have already this app this link just opens the app. I want to redirect user to that specific page when he clicks on a link.

How can I do this?

I heard something about intent-filter but not sure how to use them.

<intent-filter>
<action android:name="android.intent.action.VIEW" />
<category android:name="android.intent.category.BROWSABLE" />
<category android:name="android.intent.category.DEFAULT" />
<data android:mimeType="application/pdf" />

UPDATED

Now I am trying to share my app link like this

Intent shareIntent = new Intent(android.content.Intent.ACTION_SEND);
    shareIntent.setType("text/plain");
    shareIntent.putExtra("PostID", postID);
    shareIntent.putExtra("UserID", userID);
    shareIntent.putExtra(android.content.Intent.EXTRA_TEXT, "https://play.google.com/store/apps/details?id=" + activity.getPackageName());
    activity.startActivity(Intent.createChooser(shareIntent, "Share"));

Inside my manifest I have declared an activity like this

<activity
        android:name=".activities.ShareActivity"
        android:screenOrientation="sensorPortrait"
        android:theme="@style/Theme.AppCompat.Light.NoActionBar.FullScreen">

        <intent-filter>
            <action android:name="android.intent.action.SEND" />

            <category android:name="android.intent.category.DEFAULT" />
            <category android:name="android.intent.category.BROWSABLE" />

            <data
                android:host="details"
                android:scheme="market" />
            <data
                android:host="play.google.com"
                android:pathPattern="/store/apps/details?id=my_package_name"
                android:scheme="http" />
            <data
                android:host="play.google.com"
                android:pathPattern="/store/apps/details?id=my_package_name"
                android:scheme="https" />
        </intent-filter>
    </activity>

But I am still not able to see my app when I click on the link I shared using WhatsApp or any other means.

Please help me

Any help will be really appreciated. Thanks in advance.

回答1:

Add the following intent-filters to your activity declaration, in the AndroidManifest of your project :

    <intent-filter>

        <action android:name="android.intent.action.SEND"/>
        <category android:name="android.intent.category.DEFAULT"/>

        <data android:mimeType="text/plain"/>

    </intent-filter>

Now, in the onCreate(), onResume() or onNewIntent() method of your activity lifecycle, you can use the getData() of the intent that has fired your activity.

来源：https://stackoverflow.com/questions/35417950/how-to-open-my-application-when-app-link-is-clicked

标签

android

android-intent

intentfilter