How to launch my application from link if, it installed in device

Question

<data
  android:ssp="com.faithmusicmissions.cloudapp"
  android:path="/store/apps/details"
  android:host="play.google.com"
  android:scheme="https" />

using this code, i was try to open my application, from the play store link,
https://play.google.com/store/apps/details?id=com.faithmusicmissions.cloudapp

this code worked but the problem is when i click on other application links which contain "play.google.com" it also give the option to open my application

e.g. I have 2 play store application links in my email, from these link one of my application link and other is another application link
1. https://play.google.com/store/apps/details?id=com.faithmusicmissions.cloudapp
2. https://play.google.com/store/apps/details?id=com.tradiesfriend.tradie

when i click on both links, they display same chooser dialog to launch my application

how to use my application link in element for this https://play.google.com/store/apps/details?id=com.faithmusicmissions.cloudapp

@Hacketo thanks for giving me answer, but where i put this url in data element? is in in path? i set it like this, is it true? — Krunal Shah, Sep 22 '14 at 07:18

score 1 · Accepted Answer · answered Sep 22 '14 at 07:32

    Manifest file 

     <activity
                android:name=".SplashScreenActivity"
                android:label="@string/app_name"
                android:screenOrientation="portrait" >
                <intent-filter>
                    <action android:name="android.intent.action.MAIN" />

                    <category android:name="android.intent.category.LAUNCHER" >
                    </category>
                </intent-filter>
                <intent-filter android:label="@string/app_name" >
                    <action android:name="android.intent.action.VIEW" >
                    </action>

                    <data
                        android:host="xxx.com"
                        android:scheme="http" >
                    </data>

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

if want to access url 
In activty

try {
    Uri intentUri = getIntent().getData();
    Log.d(TAG, "Request intentUri: "+intentUri);

} catch (Exception e) {
    Log.i(TAG, "Request intentUri: " , e);
}

Thanks Sheetal Suryan, but i was implemented same code, Please see post again, i was updated and put screen shots what i am facing. — Krunal Shah, Sep 23 '14 at 05:02
You need to follow the standard rules for URIs http://stackoverflow.com/questions/2430045/how-to-register-some-url-namespace-myapp-app-start-for-accessing-your-progr — Sheetal Suryan, Sep 23 '14 at 05:57

How to launch my application from link if, it installed in device

1 Answers1