Firebase Return Child That Have Biggest Value

I am using firebase for a little while but I have stuck at one problem that I want to get the child that have the biggest value: Example here's my json (edited manually): I need t

Solution 1:

Firebase queries are ordered ascending. Since you order by rank and limit to the first result, you'll get the lowest score. To get the highest score, use limitToLast(1).

In addition, when you execute a query against the Firebase Database, there will potentially be multiple results. So the snapshot contains a list of those results. Even if there is only a single result, the snapshot will contain a list of one result.

So in total:

DatabaseReferencemDatabasePlayers= FirebaseDatabase.getInstance().getReference().child("Players");
QuerymDatabaseHighestPlayer= mDatabasePlayers.child("Scores").orderByChild("rank").limitToLast(1);
mDatabaseHighestPlayer.addValueEventListener(newValueEventListener() {
    @OverridepublicvoidonDataChange(DataSnapshot dataSnapshot){
        for (DataSnapshot childSnapshot: dataSnapshot.getChildren()) {
            StringKey= childSnapshot.getKey();
            Toast.makeText(main.this,Key,Toast.LENGTH_LONG).show();
        }
    }
    @OverridepublicvoidonCancelled(DatabaseError databaseError) {
        throw databaseError.toException(); // don't swallow errors
    }
});

Solution 2:

Change this line:

QuerymDatabaseHighestPlayer= mDatabasePlayers.child("Scores").orderByChild("rank").limitToFirst(1);

with

QuerymDatabaseHighestPlayer= mDatabasePlayers.child("Scores").child(scoreId).orderByChild("rank").limitToFirst(1);

In which scoreId is the unique id generated bu the push() method.