Java - 循环无法找到对象数组

Question

我只是制作超市程序并使用像这样的对象数组从类中初始化对象

swalayanTinyV2 sw[] = new swalayanTinyV2[3];
    sw[0] = new swalayanTinyV2("Kevin", "3812345678", 5000000, 240898 );
    sw[1] = new swalayanTinyV2("Sandra", "5612345678", 6000000, 240899);
    sw[2] = new swalayanTinyV2("Ryan", "7612345678", 7000000, 240897);

然后我进行迭代以找到与客户名称和PIN类似的对象。

int limit = 3;
for (int swalayanTinyV2 sw1 : sw) {
    while (limit!=0) {
    System.out.print("username : ");
    String username = in.nextLine();
    System.out.print("PIN      : ");
    int pin = in.nextInt();
    in.nextLine();

        if (username.equalsIgnoreCase(sw1.getCustName()) && pin==sw1.getPin())  {

        sw1.displayMessage();

        System.out.println("MENU    : ");
        System.out.println("");
        System.out.println("1. Buy Goods");
        System.out.println("2. Top Up");
        System.out.print("Input your choice : ");
        int pil = in.nextInt();
        System.out.println("");
        switch (pil) {
            case 1 :
                System.out.print("Total of price : ");
                double total = in.nextDouble();

                sw1.setTotalBelanja(total);
                sw1.getTotalBelanja();
                sw1.Struk();

                break;

            case 2 : 
                System.out.print("Add value : ");
                double topup = in.nextDouble();
                System.out.println("=============================================================");
                System.out.println("Value added  : Rp. "+topup);
                System.out.println("Your value   : Rp. "+sw1.getTopUpSaldo(topup));
                break;
                 default : System.out.println("Invalid input!");
        }


       return;
    }
        else {
        System.out.println("Incorrect username/PIN!");
        limit--;
        System.out.println("Your chance to login : "+limit);
    }
    }
        if (limit==0) {
        System.out.println("Account has been banned.");
    }
    }

但是迭代使程序只引用了数组对象的第一个索引。 （当我在数组对象的第二个索引中输入用户名和PIN时，它称为无效的用户名/ PIN）。

请帮忙。