我從sqlite獲取對象數據輸出而不是我需要的數據

Question

我從我的代碼中獲取對象數據，但我看不到我能做到這一點。

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

    Log.d(TAG, "onCreate"); 

    datasource = new DataSource(this); 
    datasource.open(); 

    List<Encouragement> values = datasource.getAllEncouragements(); 

    Log.d(TAG, "values"); 

    ArrayAdapter<Encouragement> adapter = new ArrayAdapter<Encouragement>(this, 
     android.R.layout.simple_list_item_1, values); 

    setListAdapter(adapter); 
    Log.d(TAG, "end of onCreate"); 
} 

和

public List<Encouragement> getAllEncouragements() { 

    List<Encouragement> encouragements = new ArrayList<Encouragement>(); 

    Cursor cursor = database.query(DatabaseHelper.TABLE_ENCOURAGEMENTS, 
      allColumns, null, null, null, null, null); 

    cursor.moveToFirst(); 

    while (!cursor.isAfterLast()) { 
     Encouragement encouragement = cursorToEncouragement(cursor); 
     encouragements.add(encouragement); 
     cursor.moveToNext(); 
    } 

    cursor.close(); 
    return encouragements; 
} 

是生產com.example.myActivity.myObject@34563475列表中的項目。

Answer 1

該列表視圖調用toString()來表示myObject。您尚未覆蓋toString()方法，因此調用默認的Object.toString()方法。

@Override 
public String toString(){ 
    return this.name; // whatever you need to represent this instance of myObject. 
}