如何从 Firebase (java) 中检索最后一个孩子？

Question

DatabaseReference lastRef = FirebaseDatabase.getInstance().getReference().child("Ranklist");
Query lastQuery = lastRef.orderByKey().limitToLast(1);
lastQuery.addListenerForSingleValueEvent(new ValueEventListener() {
    @Override
    public void onDataChange(DataSnapshot data) {
        newUser.Rank=Long.parseLong(data.getKey().toString())+1;

        OrderConsoleMessages();
        texts[0].setText("Last Rank is: "+newUser.Rank+" | UID: " + data.getValue());
    }

    @Override
    public void onCancelled(DatabaseError databaseError) {
        OrderConsoleMessages();
        texts[0].setText(databaseError.toString());
    }
});

My question is how do I retrieve the last child from "Ranklist"? I tried to orderByKey ByValue but nothing works in my case. I don't get any errors, I think I simply don't get any data from this query.

Edit: This is what I get from data.getValue().toString ->>> {4=somekey4}

Answer 1

your code is working correctly. I tested it. Your query to database returning the last child of the Parent Node "Ranklist". make your DatabaseReference dbref as global member

dbRef = FirebaseDatabase.getInstance().getReference().child("Ranklist");
Query lastQuery = dbRef.orderByKey().limitToLast(1);
lastQuery.addListenerForSingleValueEvent(new ValueEventListener() {
@Override
public void onDataChange(DataSnapshot dataSnapshot) {
     for(DataSnapshot data : dataSnapshot.getChildren())
     {
       //if you call methods on dataSnapshot it gives you the required values
       String s = data.getValue(); // then it has the value "somekey4" 
       String key = data.getKey(); // then it has the value "4:"
       //as per your given snapshot of firebase database data 
      }
}

@Override
public void onCancelled(DatabaseError databaseError) {

}});

Hope it is helping...

Happy Coding :-)

Answer 2

Your code is correct but you are getting only Key of the Last Node. So you need to change it with like:

dbRef.child("Ranklist")
 .orderByChild(Constants.ARG_CONV_TIMESTEMP)
 .limitToLast(1)
 .addChildEventListener(new ValueEventListener() {
    @Override
    public void onDataChange(DataSnapshot dataSnapshot) {
        String value = dataSnapshot.getValue(); 
        String key = dataSnapshot.getKey(); 
    }

    @Override
    public void onCancelled(DatabaseError databaseError) {

    }});

Answer 3

You cannot use a query in this case. As I see in your screenshot, the Ranklist node is an ArrayList and to get the value of your last child under Ranklist node, which is somekey4 , please use the following lines of code:

DatabaseReference rootRef = FirebaseDatabase.getInstance().getReference();
DatabaseReference ref = rootRef.child("Ranklist");
ValueEventListener valueEventListener = new ValueEventListener() {
    @Override
    public void onDataChange(DataSnapshot dataSnapshot) {
        ArrayList<String> arrayList = (ArrayList<String>) dataSnapshot.getValue();
        String lastItem = arrayList.get(arrayList.size() - 1);
        Log.d("TAG", lastItem);
    }

    @Override
    public void onCancelled(@NonNull DatabaseError databaseError) {
        Log.d(TAG, databaseError.getMessage());
    }
};
ref.addListenerForSingleValueEvent(valueEventListener);

The output in your logcat will be: somekey4 .