how to see the url's page when pressed to notification - parse-platform

Question

I want to send a push notification on parse.com to my Android device. Notification message will have a link inside. And when pressed, I don't want to open the application, I want to see the web page that the link is related to.
Simply, I tried the JSON below :
{
"alert" : "http://www.google.com",
"action" : "PushUrl"
}
It didn't work. Opened the application.
Is there a way to go to the link in the notification?

Answer 1

If you don't want the link to open in your application, you can invoke a browser application on clicking the notification.
Use this code to open link:
String url = "http://www.google.com";
// pending intent is redirection using the deep-link
Intent resultIntent = new Intent(Intent.ACTION_VIEW);
resultIntent.setData(Uri.parse(url));
PendingIntent pending = PendingIntent.getActivity(this, 0, resultIntent, Intent.FLAG_ACTIVITY_NEW_TASK);
notificationBuilder.setContentIntent(pending);
// using the same tag and Id causes the new notification to replace an existing one
mNotificationManager.notify(String.valueOf(System.currentTimeMillis()), 0, notificationBuilder.build());
Or create a WebView Activity in your application and open that activity upon clicking notification. This way your own app will have a browser activity and not any third party application.