Android Firebase：获取孩子内的所有对象？

Question

我有这个firebase数据库

如何获取属性项？

这是我尝试阅读的方式

for (DataSnapshot snap : dataSnapshot.getChildren()) {
                    User user = snap.getValue(User.class);

                    mPropertyList = snap.getValue(User.class).getProperty();

                    //if not current user, as we do not want to show ourselves then chat with ourselves lol
                    try {
                        mAllUserList.add(user);

                        if (mAuth.getCurrentUser().getEmail().equals(user.getEmail()))
                            if (user.getUser_type().equals("landlord")) {
                                user_rule = "landlord";
                                landUser = user;
                            }

                    } catch (Exception e) {
                        e.printStackTrace();
                    }
                }

Answer 1

要获取“属性”项，请尝试以下操作：

FirebaseUser user=FirebaseAuth.getInstance().getCurrentUser();
String userid=user.getUid();
DatabaseReference rootRef = FirebaseDatabase.getInstance().getReference();
DatabaseReference usersRef = rootRef.child("Users").child(userid).child("properties");
usersRef.addListenerForSingleValueEvent(new ValueEventListener() {
@Override
public void onDataChange(DataSnapshot dataSnapshot) {
   for(DataSnapshot datas: dataSnapshot.getChildren()){
       String desc=datas.child("description").getValue().toString();
       String type=datas.child("type").getValue().toString();
      //get other items
 }
}
@Override
public void onCancelled(DatabaseError databaseError) {
 }
});

快照在properties迭代后，您将能够获得那里的项目