出现此错误可能是在已经关闭的数据库对象上尝试进行操作。为了解决此问题，建议在每次使用完数据库后都要关闭该对象。以下是一个示例，展示了如何正确打开和关闭数据库：

public class MyDatabaseHelper extends SQLiteOpenHelper { // Database Version private static final int DATABASE_VERSION = 1; // Database Name private static final String DATABASE_NAME = "MyDatabase"; // Contacts table name private static final String TABLE_NAME = "contacts"; // Contacts Table Columns names private static final String KEY_ID = "id"; private static final String KEY_NAME = "name"; private static final String KEY_PHONE_NUMBER = "phone_number";

public MyDatabaseHelper(Context context) {
    super(context, DATABASE_NAME, null, DATABASE_VERSION);
}

@Override
public void onCreate(SQLiteDatabase db) {
    String CREATE_CONTACTS_TABLE = "CREATE TABLE " + TABLE_NAME + "("
            + KEY_ID + " INTEGER PRIMARY KEY,"
            + KEY_NAME + " TEXT,"
            + KEY_PHONE_NUMBER + " TEXT" + ")";
    db.execSQL(CREATE_CONTACTS_TABLE);
}

@Override
public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {
    // Drop older table if existed
    db.execSQL("DROP TABLE IF EXISTS " + TABLE_NAME);
    // Creating tables again
    onCreate(db);
}

public void addContact(Contact contact) {
    SQLiteDatabase db = this.getWritableDatabase();
    ContentValues values = new ContentValues();
    values.put(KEY_NAME, contact.getName());
    values.put(KEY_PHONE_NUMBER, contact.getPhoneNumber());
    // Inserting Row
    db.insert(TABLE_NAME, null, values);
    // Closing database connection
    db.close();
}

public Contact getContact(int id) {
    SQLiteDatabase db = this.getReadableDatabase();
    Cursor cursor = db.query(TABLE_NAME, new String[] { KEY_ID,
            KEY_NAME, KEY_PHONE_NUMBER }, KEY_ID + "=?",
            new String[] { String.valueOf(id) }, null, null, null, null);
    if (cursor != null)
        cursor.moveToFirst();
    Contact contact = new Contact(Integer.parseInt(cursor.getString(0)),
            cursor.getString(1), cursor.getString(2));
    // Closing database connection
    cursor.close();
    db.close();
    // return contact
    return contact;
}

}

在上面的示例中，我们可以看到每次操作后都使用了db.close()来关闭数据库对象。因此，如果您看到这个错误，请先检查是否有地方这样做了。如果没有，请添加。

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