public void lentdvd (){ /////借出DVD的类
        System.out.println("可以借出的DVD：");
        for (int w=0;w<dvd.length;w++){
            if (dvd[w]!=null){
            System.out.println(dvd[w]);
            }
        }
        System.out.println("请输入需要借出的DVD:");
        String lentname = sc.next();
        System.out.println("请输入借出的日期(以-号分隔,参照:2012-06-01):");
        lentri = sc.next(); //借出的日期
        System.out.println("请输入归还的日期(以-号分隔,参照:2012-06-01):");
        lentri1 = sc.next(); //归还的日期
        
    }

 DateFormat df = new SimpleDateFormat("yyyy-MM-dd");
 Date d1 = df.parse(lentri);
 Date d2 = df.parse(lentri1);
 long diff = d1.getTime() - d2.getTime();
 long days = diff / (1000 * 60 * 60 * 24);

------解决方案--------------------

Java code

        final double MILLS_TO_DAY = 1000d * 60 * 60 * 24;

        String sBorrow = "2011-06-01";
        String sReturn = "2012-06-01";
        
        // 可以加一些格式验证

        SimpleDateFormat formater = new SimpleDateFormat("yyyy-MM-dd");
        Date borrowDate = null;
        Date returnDate = null;
        try
        {
            borrowDate = formater.parse(sBorrow);
            returnDate = formater.parse(sReturn);
        }
        catch (ParseException e)
        {
            e.printStackTrace();
        }
        System.out.println((returnDate.getTime() - borrowDate.getTime())
                / MILLS_TO_DAY);