通过浏览器打开Android应用（如果已安装应用），否则打开playstore

Question

I have an app-url which opens app if installed and a playstore url which list my app on playstore

app-url is like custom://
and playstore url is like market://

I have to make a button in my webpage which onclicking opens app if installed otherwise open playstore.
I know there is no way in Android to detect whether the mobile has app installed or not through javascript calls.
Currently I am using intent url which does the same that i require but it does that only for chrome v 25+.
I want to do it for all browsers.
I have tried using frames and timeout but it shows a pop-up in firefox that "couldn't find an application to open the link"
What should be stable and correct way to do it for all browsers?

Answer 1

You can specify an intentfilter in the manifest file to open URL in your application.

Example is given below.

<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="http" />
      <data android:scheme="https" />
</intent-filter>

This url you could fetch from your activity using intent.