Question

我想使用Firebase Realtime Database创建一个聊天应用程序，其中包含RecyclerView和每个用户的名字的CardView。问题是，当数据库中只有3个人时，它仅向我显示Cardview。我在此链接中附加了一张数据库图片：

UsersFragment：

.Select(z => z.Value)

UserAdapter。在我中使用了User类：

public class UsersFragment extends Fragment {
    private RecyclerView recyclerView;
    private UserAdapter userAdapter;
    private List<User> mUsers;


    @Override
    public View onCreateView (LayoutInflater inflater, ViewGroup container,
                              Bundle savedInstanceState) {
        // Inflate the layout for this fragment
        View view= inflater.inflate ( R.layout.fragment_users3, container, false );
recyclerView=view.findViewById (R.id.recycler_view  );

recyclerView.setLayoutManager ( new LinearLayoutManager ( getContext () ) );
mUsers=new ArrayList<> (  );
readUsers();

        return view;
    }

    private void readUsers () {
        final FirebaseUser firebaseUser= FirebaseAuth.getInstance ().getCurrentUser ();
        DatabaseReference referenc= FirebaseDatabase.getInstance ().getReference ("users");

        referenc.addValueEventListener ( new ValueEventListener () {
            @Override
            public void onDataChange (@NonNull DataSnapshot dataSnapshot) {
                mUsers.clear ();
                for(DataSnapshot snapshot:dataSnapshot.getChildren ()){
                    User user=snapshot.getValue (User.class);
                    assert user!=null;
                    assert firebaseUser!=null;
                    if(!firebaseUser.getUid ().equals ( user.getUid () )){
                        mUsers.add ( user );

                    }
                }
                userAdapter=new UserAdapter ( getContext (),mUsers );
                recyclerView.setAdapter ( userAdapter );
            }

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

            }
        } );
    }


}

存储名字的用户：

public class UserAdapter extends RecyclerView.Adapter<UserAdapter.ViewHolder> {
    private Context mContext;
    private List<User> mUsers;
    public UserAdapter(Context mContext,List<User> mUsers){
        this.mUsers=mUsers;
        this.mContext=mContext;
    }

    @NonNull
    @Override
    public ViewHolder onCreateViewHolder (@NonNull ViewGroup parent, int viewType) {
        View view= LayoutInflater.from ( mContext ).inflate ( R.layout.users_template ,parent,false);

        return new UserAdapter.ViewHolder ( view );
    }

    @Override
    public void onBindViewHolder (@NonNull ViewHolder holder, int position) {
User user=mUsers.get ( position );
holder.username.setText ( user.getFirstName () );

    }

    @Override
    public int getItemCount () {
        return mUsers.size ();
    }

    public class ViewHolder extends RecyclerView.ViewHolder{

        public TextView username;

        public ViewHolder(View itemView){
            super(itemView);
            username=itemView.findViewById ( R.id.username1 );
        }

    }
}

还有我的TabLayout和ViewPager所在的活动

public class User {
   public String firstName;
    public String secondName;
    public String uid;
    public String email;
    public User(){

    }



    public User (String firstName, String secondName, String uid, String email) {
        this.firstName = firstName;
        this.secondName = secondName;
        this.uid=uid;
        this.email=email;
    }

    public User (String firstname, String secondname) {
    }

    public String getFirstName () {
        return firstName;
    }

    public String getSecondName () {
        return secondName;
    }

    public String getUid () {
        return uid;
    }

    public String getEmail () {
        return email;
    }

    public void setFirstName (String firstName) {
        this.firstName = firstName;
    }

    public void setSecondName (String secondName) {
        this.secondName = secondName;
    }

    public void setUid (String uid) {
        this.uid = uid;
    }

    public void setEmail (String email) {
        this.email = email;
    }
}

在调试模式下，它显示给我：

public class UsersActivity extends AppCompatActivity {


    @Override
    protected void onCreate (Bundle savedInstanceState) {
        super.onCreate ( savedInstanceState );
        setContentView ( R.layout.activity_users );
        TabLayout tablayout=findViewById ( R.id.tab_layout );
        ViewPager viewPager=findViewById ( R.id.view_pager );
        ViewPagerAdapter viewPagerAdapter=new ViewPagerAdapter ( getSupportFragmentManager () );
        viewPagerAdapter.addFragment ( new ChatFragment (),"Chats" );
        viewPagerAdapter.addFragment ( new UsersFragment (),"Users" );
        viewPager.setAdapter (viewPagerAdapter  );
        tablayout.setupWithViewPager (viewPager  );

    }
    class ViewPagerAdapter extends FragmentPagerAdapter{
private ArrayList<Fragment> fragments;
private ArrayList<String> titles;
ViewPagerAdapter(FragmentManager fn){
    super(fn);
    this.fragments=new ArrayList<> (  );
    this.titles=new ArrayList<> (  );
}
        @NonNull
        @Override
        public Fragment getItem (int position) {
            return fragments.get ( position );
        }

        @Override
        public int getCount () {
            return fragments.size ();
        }
        public void addFragment(Fragment fragment,String title){
    fragments.add ( fragment );
    titles.add ( title );
        }

        @Nullable
        @Override
        public CharSequence getPageTitle (int position) {
            return titles.get ( position );
        }
    }
}

我希望向我展示我在数据库中拥有的所有用户。

Answer 1

将Firebase数据库中的密钥更改为firstName，secondName，email，而不是FirstName，SecondName和E-mail。如果在模型类中找不到正确的键及其getter和setter，则无法映射firebase中的数据。

Answer 2

您在数据库中的JSON中具有属性FirstName。在User类中，有一个方法getFirstName()，该方法对应于属性firstName（根据Firebase使用的JavaBean约定）。由于情况不同，因此属性不同。

要使JSON和类匹配，可以将JSON更改为使用JavaBean属性名称：

{
  "users": {
    "uid1": {
      "email": "...",
      "firstName": "...",
      "secondName": "..."
    }
  }
}

或者，您可以使用PropertyName属性使Java类与当前JSON匹配：

public class User {
    private String firstName;
    private String secondName;
    private String uid;
    private String email;

    public User(){
    }

    public User (String firstName, String secondName, String uid, String email) {
        this.firstName = firstName;
        this.secondName = secondName;
        this.uid=uid;
        this.email=email;
    }

    public User (String firstname, String secondname) {
    }

    @PropertyName("FirstName")
    public String getFirstName () {
        return firstName;
    }

    @PropertyName("SecondName")
    public String getSecondName () {
        return secondName;
    }

    @PropertyName("Uid")
    public String getUid () {
        return uid;
    }

    @PropertyName("E-mail")
    public String getEmail () {
        return email;
    }

    @PropertyName("FirstName")
    public void setFirstName (String firstName) {
        this.firstName = firstName;
    }

    @PropertyName("SecondName")
    public void setSecondName (String secondName) {
        this.secondName = secondName;
    }

    @PropertyName("Uid")
    public void setUid (String uid) {
        this.uid = uid;
    }

    @PropertyName("E-mail")
    public void setEmail (String email) {
        this.email = email;
    }
}

也请在这里查看我的答案：Specifying serialized names in java class for firestore document

用户列表为空

2 个答案: