从 Firebase 获取项目的位置

Question

我找到了通过我的排行榜列表运行并返回相应值的代码。 但我需要返回列表中项目的位置。 我应该如何修改代码以便它返回项目的位置（索引）。

谢谢。

//query for sorting by score
    score = FirebaseDatabase.getInstance().getReference().child("Leaders");
    final Query query = score.orderByChild("uID").equalTo(mCurrentUser.getUid());

    query.addValueEventListener(new ValueEventListener() {
        @Override
        public void onDataChange(DataSnapshot dataSnapshot) {

            Iterator<DataSnapshot> iterator = dataSnapshot.getChildren().iterator();

            int length = (int) dataSnapshot.getChildrenCount();

            String[] sampleString = new String[length];

            while(i < length) {
                sampleString[i] = iterator.next().getValue().toString();
                //Log.d(Integer.toString(i), sampleString[i]);

                //print the value of current score
                Toast.makeText(LeaderBoardActivity.this,sampleString[i],Toast.LENGTH_LONG).show();
                i++;
            }
        }

        @Override
        public void onCancelled(DatabaseError databaseError) {

        }
    });

Answer 1

我已经根据您的评论编辑了代码。 既然你只关心位置和分数...

// you will need to query the database based on scores
score = FirebaseDatabase.getInstance().getReference().child("Leaders");
final Query query = score.orderByChild("Score");
query.addListenerForSingleValueEvent(new ValueEventListener() {
    @Override
    public void onDataChange(DataSnapshot dataSnapshot) {
        // get total leaders. Let's say total leader is 5 persons
        int total = (int) dataSnapshot.getChildrenCount();
        // let's say userB is actually 2nd in the list
        int i = 0;
        // loop through dataSnapshot
        for (DataSnapshot childSnapshot : dataSnapshot.getChildren()) {
            String uID = childSnapshot.child("uID").getValue();
            if (uID.equals(mCurrentUser.getUid()) {
                // the list is ascending, when finally reach userB, i = 3
                int userPlace = total - i;    // gives 2
                int score = childSnapshot.child("Score").getValue(Interger.class);
                break;
            } else {
                i++;
            }
        }
    }

    @Override
    public void onCancelled(DatabaseError databaseError) {

    }
});

请注意，这可能非常消耗带宽。 随着用户越来越多，您将需要下载每个查询的所有用户数据。 如果你只展示前 5 名领导者等可能会更好。

Answer 2

我使用以下方法按位置更改子值，希望这可以给您一些帮助：

DataSnapshot targetData;

@Override
protected void onStart() {
    super.onStart();

    targetReference.addValueEventListener(new ValueEventListener() {
        @Override
        public void onDataChange(@NonNull DataSnapshot dataSnapshots) {
            targetData = dataSnapshots;        //save in global variable
        }

        @Override
        public void onCancelled(@NonNull DatabaseError databaseError) {
        }
    });
}

private void yourMethod() {
    int index = 0;

    for (DataSnapshot childSnapshot : targetData.getChildren()) {
        if (index == 1) {        //your target index
            DatabaseReference currentReference = childSnapshot.getRef();

            currentReference.setValue("xxx");
        }

        index++;
    }
}