Firebase retrieve highest 100 score

Question

This is a screen shot of my firebase: I am trying to retrieve the highest 100 score in firebase database I am using this code to add new node to firebase:

 Map<String, String> post1 = new HashMap<String, String>();
        post1.put("name",name);
        post1.put("score",score);
        myRef.push().setValue(post1);

And this is the code I am using to retrieve the highest 100 score which doesn't work (the code works but it is not retrieving the highest 100 score)

 Query queryRef = myRef.orderByChild("score").limitToFirst(100);
        queryRef.addValueEventListener(new ValueEventListener() {
            @Override
            public void onDataChange(DataSnapshot dataSnapshot) {

                for (DataSnapshot postSnapshot: dataSnapshot.getChildren()) {
                    Score score=postSnapshot.getValue(Score.class);
                    Log.d("test"," values is " + score.getName()  + " " + score.getScore());
                }
            }

            @Override
            public void onCancelled(DatabaseError databaseError) {

            }
        }); 

Answer 1

Firebase queries are always in ascending order. So you'll need to get the last 100, instead of the first 100.

Query queryRef = myRef.orderByChild("score").limitToLast(100);

Then client-side you'll need to reverse the items.

Alternatively you can add a inverted property to your items invertedScore: -99. If you do that, you can order by that inverted score and won't have to reverse the array.

This scenario has been covered frequently before. I highly recommend you study some of these:

• data sorting in firebase
• Firebase Leaderboard, best Implmentation. SQL Joins and orderby (as recently as yesterday)
• Display posts in descending posted order
• Firebase Data Desc Sorting in Android
• Swift - How to create Sort query as Descending on Firebase?