java - SQLite db 重复值 Android

Question

我在应用程序中使用 spinner 从 SQLite 数据库获取值。问题是每次我从 Eclipse 运行应用程序时，值都会一次又一次加载。因此，如果第一次我在 spinner 中有 5 个值，第二次我有 10 个值，然后是 15，然后是 20，等等。我该如何解决这个问题？

这是数据库部分:

在onCreate中:

createTable();
        insertIntoTable();

        ArrayList<String> my_array = new ArrayList<String>();
        my_array = getTableValues();

        My_spinner = (Spinner) findViewById(R.id.scelte);
        My_spinner.setOnItemSelectedListener(this);
        ArrayAdapter my_Adapter = new ArrayAdapter(this, R.layout.spinner_row, my_array);
        My_spinner.setAdapter(my_Adapter);

然后:

// CREATE TABLE IF NOT EXISTS
    public void createTable() {
        try {
            mydb = openOrCreateDatabase(DBNAME, Context.MODE_PRIVATE, null);
            mydb.execSQL("CREATE TABLE IF  NOT EXISTS " + TABLE
                    + " (ID INTEGER PRIMARY KEY, NAME TEXT, PLACE TEXT);");
            mydb.close();
        } catch (Exception e) {
            Toast.makeText(getApplicationContext(), "Errore durante la creazione della tabella",
                    Toast.LENGTH_LONG);
        }
    }

    // THIS FUNCTION INSERTS DATA TO THE DATABASE
    public void insertIntoTable() {
        try {
            mydb = openOrCreateDatabase(DBNAME, Context.MODE_PRIVATE, null);
            mydb.execSQL("INSERT INTO " + TABLE
                    + "(NAME, PLACE) VALUES('Cibo e Bevande','cibo')");
            mydb.execSQL("INSERT INTO " + TABLE
                    + "(NAME, PLACE) VALUES('Trasporti','trasporti')");
            mydb.execSQL("INSERT INTO " + TABLE
                    + "(NAME, PLACE) VALUES('Svago','svago')");
            mydb.execSQL("INSERT INTO " + TABLE
                    + "(NAME, PLACE) VALUES('Acquisti vari','acquisti')");
            mydb.execSQL("INSERT INTO " + TABLE
                    + "(NAME, PLACE) VALUES('Sanità','sanità')");
            mydb.execSQL("INSERT INTO " + TABLE
                    + "(NAME, PLACE) VALUES('Altro','altro')");
            mydb.close();


        } catch (Exception e) {
            Toast.makeText(getApplicationContext(),
                    "Error in inserting into table", Toast.LENGTH_LONG);
        }
    }

    // THIS FUNCTION SHOWS DATA FROM THE DATABASE
    public ArrayList<String> getTableValues() {

        ArrayList<String> my_array = new ArrayList<String>();
        try {
            mydb = openOrCreateDatabase(DBNAME, Context.MODE_PRIVATE, null);
            Cursor allrows = mydb.rawQuery("SELECT * FROM " + TABLE, null);
            System.out.println("COUNT : " + allrows.getCount());

            if (allrows.moveToFirst()) {
                do {

                    String ID = allrows.getString(0);
                    String NAME = allrows.getString(1);
                    String PLACE = allrows.getString(2);
                    my_array.add(NAME);


                } while (allrows.moveToNext());
            }
            allrows.close();
            mydb.close();
        } catch (Exception e) {
            Toast.makeText(getApplicationContext(), "Errore.",
                    Toast.LENGTH_LONG);
        }
        return my_array;

    }

    @Override
    public void onItemSelected(AdapterView<?> arg0, View arg1, int arg2,
            long arg3) {
        // TODO Auto-generated method stub
        String selecteditem = My_spinner.getSelectedItem().toString();
        TextView categorietext = (TextView) findViewById(R.id.sceltelabel);
        categorietext.setText(selecteditem);
    }

    @Override
    public void onNothingSelected(AdapterView<?> arg0) {
        // TODO Auto-generated method stub

    }

Answer 1

你运行这个

insertIntoTable();

每次在onCreate中都会添加越来越多的数据。

您应该使用扩展SQLiteOpenHelper类并使用onCreate方法来创建数据库并初始化表。请参阅Android developer reference .