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我试图查询当前用户的好友列表并将其添加到 android 中的 listview 中。问题是方法不断显示错误

这是我的代码

package com.example.krisanapongpoonsawat.chatit;

import android.app.ProgressDialog;
import android.content.Intent;
import android.os.Bundle;
import android.support.design.widget.FloatingActionButton;
import android.support.design.widget.Snackbar;
import android.support.v7.app.AppCompatActivity;
import android.support.v7.widget.Toolbar;
import android.util.Log;
import android.view.Menu;
import android.view.MenuInflater;
import com.parse.FindCallback;
import android.view.MenuItem;
import android.view.View;
import android.widget.AdapterView;
import android.widget.ArrayAdapter;
import android.widget.Button;
import android.widget.ListView;
import android.widget.Toast;

import com.parse.FindCallback;
import com.parse.Parse;
import com.parse.ParseException;
import com.parse.ParseObject;
import com.parse.ParseQuery;
import com.parse.ParseRelation;
import com.parse.ParseUser;

import java.util.ArrayList;
import java.util.List;

public class User extends AppCompatActivity {
    private ArrayList<ParseUser> uList;

    /** The user. */
    public static ParseUser user;
    ArrayList<String> friendlist;
    ArrayAdapter<String> listAdapter;
    ListView list;

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.user);
        Toolbar myToolbar = (Toolbar) findViewById(R.id.toolbar);
        setSupportActionBar(myToolbar);
        final ParseUser currentUser = ParseUser.getCurrentUser();
        list = (ListView) findViewById(R.id.listView);
        listAdapter = new ArrayAdapter<String>(this,
                android.R.layout.simple_list_item_1);
        list.setAdapter(listAdapter);

    }
    @Override
    protected void onResume()
    {
        super.onResume();
        loadUserList();

    }
    private void loadUserList()
    {
        ParseQuery query = new ParseQuery("Friends");
        query.whereEqualTo("owner", ParseUser.getCurrentUser().getObjectId().toString());
        query.findInBackground(new FindCallback() {
            public void done(List<ParseObject> friendList, ParseException e) {
                if (e == null) {
                    Log.d("score", "Retrieved " + friendList.size() + " scores");
                    for (int i = 0; i < friendlist.size(); i++) {
                        Object object = friendlist.get(i);
                        String name = ((ParseObject) object).getObjectId().toString();
                        listAdapter.add(name);


                    }
                } else {
                    Log.d("score", "Error: " + e.getMessage());
                }
            }
        });


    }


    @Override
    public boolean onCreateOptionsMenu(Menu menu) {
        MenuInflater inflater = getMenuInflater();
        inflater.inflate(R.menu.menu, menu);
        return true;
    }
    @Override
    public boolean onOptionsItemSelected(MenuItem item) {
        switch (item.getItemId()) {
            case R.id.action_favorite:
                // User chose the "Settings" item, show the app settings UI...
                Intent intent = new Intent(User.this, Search.class);
                startActivity(intent);
                finish();
                return true;

            default:
                // If we got here, the user's action was not recognized.
                // Invoke the superclass to handle it.
                return super.onOptionsItemSelected(item);

        }
    }




}

并且显示的错误是

    Error:(66, 51) error: <anonymous com.example.krisanapongpoonsawat.chatit.User$1> is not abstract and does not override abstract method done(List,ParseException) in FindCallback
Error:(67, 25) error: name clash: done(List<ParseObject>,ParseException) in <anonymous com.example.krisanapongpoonsawat.chatit.User$1> and done(List<T>,ParseException) in FindCallback have the same erasure, yet neither overrides the other
where T is a type-variable:
T extends ParseObject declared in interface FindCallback

你能解决这个问题吗？我已经坚持了3天了

如果您需要任何其他信息，请告诉我！

谢谢

4

1 回答 1

0

这里有很多关于关系的信息

参考本文档和我们在评论中的对话，我认为您的查询应该如下所示：

ParseQuery<ParseObject> query = ParseQuery.getQuery("_User");
query.whereEqualTo("owner", ParseUser.getCurrentUser().getObjectId().toString());
query.findInBackground(new FindCallback<ParseObject>() {
    public void done(List<ParseObject> friendList, ParseException e) {
        if (e == null) {
            Log.d("score", "Retrieved " + friendList.size() + " scores");
            for (int i = 0; i < friendList.size(); i++) {
                ParseObject item = friendList.get(i);
                ParseRelation<ParseObject> friendRelation = item.getRelation("Friends");
                ParseQuery friendsQuery = friendRelation.getQuery();
                findFriends(friendsQuery);
            }
         } else {
            Log.d("score", "Error: " + e.getMessage());
         }
    }
});

...

private void findFriends(ParseQuery query) {

    query.findInBackground(new FindCallback() {

        ...

    });

}

我的回答中的某些内容很可能不起作用，但我没有其他想法。

于 2015-11-21T21:37:45.913 回答