如何在java中找到两个日期之间的差异持续时间？

Question

如何在java中找到两个日期之间的差异持续时间？

我有两个DateTime对象，需要找到它们的差异持续时间，

我有以下代码，但不知道如何继续它以获得预期的结果如下：

例

      11/03/14 09:30:58

      11/03/14 09:33:43

      elapsed time is 02 minutes and 45 seconds

      -----------------------------------------------------

      11/03/14 09:30:58 

      11/03/15 09:30:58

      elapsed time is a day

      -----------------------------------------------------

      11/03/14 09:30:58 

      11/03/16 09:30:58

      elapsed time is two days

      -----------------------------------------------------

      11/03/14 09:30:58 

      11/03/16 09:35:58

      elapsed time is two days and 05 mintues

码

    String dateStart = "11/03/14 09:29:58";

    String dateStop = "11/03/14 09:33:43";

    Custom date format

    SimpleDateFormat format = new SimpleDateFormat("yy/MM/dd HH:mm:ss");

    Date d1 = null;

    Date d2 = null;

    try {

        d1 = format.parse(dateStart);

        d2 = format.parse(dateStop);

    } catch (ParseException e) {

        e.printStackTrace();

    }

    // Get msec from each, and subtract.

    long diff = d2.getTime() - d1.getTime();

    long diffSeconds = diff / 1000 % 60;

    long diffMinutes = diff / (60 * 1000) % 60;

    long diffHours = diff / (60 * 60 * 1000);

    System.out.println("Time in seconds: " + diffSeconds + " seconds.");

    System.out.println("Time in minutes: " + diffMinutes + " minutes.");

    System.out.println("Time in hours: " + diffHours + " hours.");

Answer 1

尝试以下方法{         Date dt2 = new DateAndTime().getCurrentDateTime();         long diff = dt2.getTime() - dt1.getTime();         long diffSeconds = diff / 1000 % 60;         long diffMinutes = diff / (60 * 1000) % 60;         long diffHours = diff / (60 * 60 * 1000);         int diffInDays = (int) ((dt2.getTime() - dt1.getTime()) / (1000 * 60 * 60 * 24));         if (diffInDays > 1) {             System.err.println("Difference in number of days (2) : " + diffInDays);             return false;         } else if (diffHours > 24) {             System.err.println(">24");             return false;         } else if ((diffHours == 24) && (diffMinutes >= 1)) {             System.err.println("minutes");             return false;         }         return true;}

Answer 2

可以使用Java内置类TimeUnit以更好的方式处理日期差异转换。它提供了实用方法：Date startDate = // Set start dateDate endDate   = // Set end datelong duration  = endDate.getTime() - startDate.getTime();long diffInSeconds = TimeUnit.MILLISECONDS.toSeconds(duration);long diffInMinutes = TimeUnit.MILLISECONDS.toMinutes(duration);long diffInHours = TimeUnit.MILLISECONDS.toHours(duration);long diffInDays = TimeUnit.MILLISECONDS.toDays(duration);

Answer 3

使用Joda-Time库DateTime startTime, endTime;Period p = new Period(startTime, endTime);long hours = p.getHours();long minutes = p.getMinutes();Joda Time有一个时间间隔的概念：Interval interval = new Interval(oldTime, new Instant());另一个例子 日期差异还有一个链接或者使用Java-8（集成了Joda-Time概念）Instant start, end;//Duration dur = Duration.between(start, stop);long hours = dur.toHours();long minutes = dur.toMinutes();