求时间差解决思路

求时间差

Java code

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            }

Java code

 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);
[解决办法]
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);