java - 如何在 Firebase 数据库 Android 上删除选定的子项

Question

我试图通过将唯一键设置为空来从数据库中删除子项。我所做的是 - 按住列表项以获取警报对话框，然后按 删除我选择的条目。问题是当我检索 key 时它会返回所有 key 。到目前为止我最接近的是从数据库中删除所有内容。请检查下面我的代码快照:

 mLocationListView.setLongClickable(true);
    mLocationListView.setOnItemLongClickListener(new AdapterView.OnItemLongClickListener() {
        @Override
        public boolean onItemLongClick(AdapterView<?> parent, View view, int position, long id) {

            AlertDialog.Builder builder = new AlertDialog.Builder(MainActivity.this)
                    .setTitle("Delete a Location")
                    .setMessage("Do you want to delete your location?")
                    .setPositiveButton("Delete", new DialogInterface.OnClickListener() {
                        @Override
                        public void onClick(DialogInterface dialog, int which) {

//Problem is here. if I exclude .child(pushKey)everything is deleted from the database. 
                                DatabaseReference db_node = FirebaseDatabase.getInstance().getReference().getRoot().child("locations").child(pushKey);

                        db_node.setValue(null);
                        }
                    });
                AlertDialog alert = builder.create();
                alert.show();
            return true;
        }
    });

这是我检索pushKey的方法。有什么建议如何以更正确的方式做到这一点？ (如果有的话)

 queryRef.addChildEventListener(new ChildEventListener() {
        @Override
        public void onChildAdded(DataSnapshot dataSnapshot, String s) {
            locationCurrent = dataSnapshot.getValue(LocationCurrent.class);
            mLocationAdapter.add(locationCurrent);
            pushKey = dataSnapshot.getKey();
            Log.i("PUSH KEY ", pushKey);

        }

        @Override
        public void onChildChanged(DataSnapshot dataSnapshot, String s) {
            locationCurrent = dataSnapshot.getValue(LocationCurrent.class);


        }

        @Override
        public void onChildRemoved(DataSnapshot dataSnapshot) {

        }

        @Override
        public void onChildMoved(DataSnapshot dataSnapshot, String s) {

        }

        @Override
        public void onCancelled(DatabaseError databaseError) {

        }
    });

我正在考虑将 key 也放入数据库，但没有意义，必须有最有效的解决方案。

更新:

我尝试使用另一种方法来执行此操作，将单值事件监听器附加到查询，但它不起作用，因为我无法确定单击哪个列表项，因此我删除了整个数据库。基本上，我需要单击列表项并将其删除。但它不起作用。我完全按照下面的链接进行操作:

How to delete from firebase realtime database?

有什么建议吗？

Answer 1

在 LocationCurrent.class 中添加一个名为 refKey 的 transient 变量。将引用保存在 onChildAdded

中

public void onChildAdded(DataSnapshot dataSnapshot, String s) {
    locationCurrent = dataSnapshot.getValue(LocationCurrent.class);
    locationCurrent.setRefKey(dataSnapshot.getKey()); 
    mLocationAdapter.add(locationCurrent);
}

然后在删除时单击执行以下操作:

mLocationListView.setOnItemLongClickListener(new AdapterView.OnItemLongClickListener() {
    @Override
    public boolean onItemLongClick(AdapterView<?> parent, View view, int position, long id) {
    DatabaseReference db_node = FirebaseDatabase.getInstance().getReference().child("locations").child(mLocationAdapter.getItem(position).getRefKey());
    db_node.removeValue();
});