Question

我正在尝试使用SimpleAdapter和TableView显示我的SQLite表的内容。它在首次启动时工作正常，但是当我重新启动Activity时，它会再次在listView中添加表。每次我从我的前菜单启动意图时都会发生这种情况。

在我看来，SimpleAdapter是从上次存储的，并且当从菜单中打开intent时再次添加？

也许有一种方法可以在每次启动时重置适配器？

我是编程新手并希望有人可以帮助我:)。

一些代码：

static final ArrayList<HashMap<String,String>> results = new ArrayList<HashMap<String,String>>(); 
private String tableName = DBAdapter.tableName;
private SQLiteDatabase newDB;



@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    openAndQueryDatabase();

    displayResultList();

    }



private void displayResultList() {

    ListAdapter adapter = new SimpleAdapter(
            this,
            results,
            R.layout.liste,
            new String[] {"Name","LastName","Age"},
            new int[] {R.id.item_id,R.id.item_name,R.id.item_tid}
            );
    setListAdapter(adapter);

}
    private void openAndQueryDatabase() {
        try {
            DBAdapter dbHelper = new DBAdapter(this.getApplicationContext());
            newDB = dbHelper.getWritableDatabase();
            Cursor c = newDB.rawQuery("SELECT Startnr, Navn, Tid FROM " +
                    tableName + " ORDER BY Startnr ASC" , null);

            if (c != null ) {
                if  (c.moveToFirst()) { 
                    do {
                        HashMap<String,String> list = new HashMap<String,String>();
                        String firstName = c.getString(c.getColumnIndex("Startnr"));
                        String lastName = c.getString(c.getColumnIndex("Navn"));
                        String age = c.getString(c.getColumnIndex("Tid"));
                        list.put("Name", "" + firstName);
                        list.put("LastName", "" + lastName);
                        list.put("Age", "" + age);
                        results.add(list);
                    }while (c.moveToNext());
                } 
            }           
        } catch (SQLiteException se ) {
            Log.e(getClass().getSimpleName(), "Could not create or Open the database");
        } finally {
            if (newDB != null) 
                newDB.close();                  
        }

    }

Answer 1

Its because you have declared the "results" array list as a "static" variable.

每次启动活动时，都要避免在listview中添加SQLite表

1 个答案: