使用 FirebaseListAdapter 显示用户列表

Question

我正在尝试将 "Search User" Activity 添加到我正在尝试制作的简单应用程序中。但是我无法获取用户输入（来自 SearchView）并使用它查询 Firebase 数据库以最终在 Listview 中显示所有内容（使用 FirebaseListAdapter）。

问题主要出在我不太明白的 Firebase 部分。这是代码：

SearchView userSearchView;
ListView userListView;
SearchUserListAdapter UserAdapter;
String currentUserName;
String currentEnteredQuery = null;


@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_search_user);

    userSearchView = (SearchView) findViewById(R.id.user_searchview);
    userListView = ( ListView) findViewById(R.id.listview_search);
    final Intent startConversation = new Intent(this,ConversationActivity.class);

    //Start a new conversation
    userListView.setOnItemClickListener(new AdapterView.OnItemClickListener() {
        @Override
        public void onItemClick(AdapterView<?> parent, View view, int position, long id) {
            startActivity(startConversation);
        }
    });

    Query query = UserDbRef.orderByChild(currentEnteredQuery);
    final FirebaseListAdapter firebaseListAdapter = new FirebaseListAdapter<UserItem>(this, UserItem.class, R.layout.user_item, query) {
        @Override
        protected void populateView(View v, UserItem userItem) {
            ((TextView)v.findViewById(android.R.id.text1)).setText(userItem.getUser_name());
        }
    };
    userListView.setAdapter(firebaseListAdapter);


    userSearchView.setOnQueryTextListener(new SearchView.OnQueryTextListener() {

        @Override
        public boolean onQueryTextSubmit(String inputQuery) {
            currentEnteredQuery = inputQuery;
            return true;
        }

        @Override
        public boolean onQueryTextChange(String newText) {
            return false;
        }
    });


}

@Override
public boolean onCreateOptionsMenu(Menu menu) {
    // Inflate the menu; this adds items to the action bar if it is present.
    getMenuInflater().inflate(R.menu.menu_search_user, menu);
    return true;
}

Answer 1

将您的查询放在 SearchView 的侦听器中，以便每次您使用特定条件搜索时它都会查询：

userSearchView.setOnQueryTextListener(new SearchView.OnQueryTextListener() {

    @Override
    public boolean onQueryTextSubmit(String inputQuery) {
        currentEnteredQuery = inputQuery;
        //Your Firebase Query here
        Query query = UserDbRef.orderByChild(currentEnteredQuery);
        final FirebaseListAdapter firebaseListAdapter = new FirebaseListAdapter<UserItem>(this, UserItem.class, R.layout.user_item, query) {
            @Override
            protected void populateView(View v, UserItem userItem) {
                ((TextView)v.findViewById(android.R.id.text1)).setText(userItem.getUser_name());
            }
        };
        userListView.setAdapter(firebaseListAdapter);
        return true;
    }

    @Override
    public boolean onQueryTextChange(String newText) {
        return false;
    }
});

还有这些行：

final FirebaseListAdapter firebaseListAdapter = new FirebaseListAdapter<UserItem>(this, UserItem.class, R.layout.user_item, query) {
    @Override
    protected void populateView(View v, UserItem userItem) {
        ((TextView)v.findViewById(android.R.id.text1)).setText(userItem.getUser_name());
    }
};

您正在使用 R.layout.user_item 作为列表的布局，但您正在使用 android.R.id.text1 .据我所知，android.R是AndroidSDK内置的资源。如果您使用 android.R.layout.simple_list_item_1，我认为这就是视图。继续检查你的 R.layout.user_item 并将 id 更改为布局中的相应视图。

使用 FirebaseListAdapter 显示用户列表

Using FirebaseListAdapter to display a list of User

android

firebase

firebase-realtime-database

firebaseui