用户列表为空 [英] List of users is empty

问题描述

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. 我在此链接中附有一张数据库图片:

I want to create a chat app using Firebase Realtime Database with a RecyclerView and a CardView with the first name for every user.The problem is that it's only show me a cardview when I have 3 people in the database. I attached in this link a picture with my database:

UsersFragment:

The UsersFragment:

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) {

            }
        } );
    }


}

UserAdapter.在我中使用了User类:

The UserAdapter.In i used the User class:

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 );
        }

    }
}

存储名字的用户:

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;
    }
}

还有我的TabLayout和ViewPager所在的活动

And the activity where my TabLayout and ViewPager is

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 );
        }
    }
}

在调试模式下，它显示给我:

In the debug mode it show me:

W/ClassMapper: No setter/field for FirstName found on class com.example.sportsbuddy.User (fields/setters are case sensitive!)

我希望向我展示我在数据库中拥有的所有用户.

I expect to show me all the users that I have in the database.

推荐答案

您在数据库中的JSON中具有属性FirstName.在User类中，您具有方法getFirstName()，该方法对应于(根据Firebase使用的JavaBean约定)属性firstName.由于情况不同，因此是不同的属性.

You have a property FirstName in your JSON in the database. In your User class, you have a method getFirstName(), which corresponds (according to the JavaBean convention that Firebase uses) to a property firstName. Since the case is different, that is a different property.

要使JSON和类匹配，可以将JSON更改为使用JavaBean属性名称:

To make the JSON and the class match, you can change the JSON to use JavaBean property names:

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

或者，您可以使用PropertyName属性使Java类与您当前的JSON相匹配:

Alternatively you can make the Java class match your current JSON by using the PropertyName attribute:

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;
    }
}

也请在此处查看我的答案:在Java中指定序列化名称Firestore文档的类

Also see my answer here: Specifying serialized names in java class for firestore document

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