如何从stringView中的listview中保存checkbox id以在查询SQLite android中进行比较？

Question

我在我的Android应用程序中有listview with checkbox。我想保存arraylist<string>中已选中复选框的ID。我想用它在我的查询中进行比较，但我在getcount()中有错误，因为查询没有运行。

这里是我的代码＆＃34; Main.java＆＃34;：

public void clicksubmit() {
    ArrayList<String> lst = new ArrayList<String>();
    lst.clear();
    String arrId = "";
    System.out.println("check"+getListView().getCheckedItemIds().length);;

    for (int i = 0; i < list.size(); i++) {
        if (list.get(i).isCek()) {
            arrId = arrId  + list.get(i).getIdGejala() +",";
        }
    }

    arrId = arrId.substring(0, arrId.length() - 1);

    Intent intent = new Intent(this, HasilDiagnosis.class);
    intent.putExtra("arrId", arrId);

    startActivity(intent);  
    Toast.makeText(this, "checked : " + arrId, Toast.LENGTH_LONG).show();
}

这是我的DBHandler.java：

    public String[][] LihatHasil(String arrId){
    try {
        String arrData[][] = null;
        SQLiteDatabase db = this.getWritableDatabase();
        String strSQL = "SELECT p.*, COUNT( idRule ) as kejadian, "
                + "(SELECT COUNT( idRule ) kejadian FROM  tbrule WHERE idGangguan=p.idGangguan) as peluang "
                + "FROM  tbrule r JOIN tbgangguan p ON p.idGangguan = r.idGangguan "
                + "WHERE r.idGejala IN ("+arrId+")"
                + "ORDER BY kejadian DESC LIMIT 3 "
                + "GROUP BY r.idPenyakit"; 
        Cursor cursor = db.rawQuery( strSQL, null);

        if (cursor.moveToFirst()) {
            arrData = new String[cursor.getCount()][cursor.getColumnCount()];
            int i= 0;
            do {
                arrData[i][0] = cursor.getString(0);
                arrData[i][1] = cursor.getString(1);
                arrData[i][2] = cursor.getString(2);
                arrData[i][3] = cursor.getString(3);
                arrData[i][4] = cursor.getString(4);
                i++;
            } while (cursor.moveToNext());
        }
        cursor.close();
        db.close();
        return arrData;

    } catch (Exception e) {
        return null;
    }

有人可以帮助我。抱歉我的英文不好