如何在java中找到两个日期之间的差异持续时间？
如何在java中找到两个日期之间的差异持续时间？
提问于 2013-07-30 14:55:12
我有两个DateTime对象，它们需要找到它们的difference的持续时间，
我有以下代码，但不确定如何继续它才能达到预期的结果，如下所示：
示例
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 minutes代码
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.");回答 18
Stack Overflow用户
回答已采纳
发布于 2013-11-10 14:58:45
请尝试以下操作
{
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;
}Stack Overflow用户
发布于 2013-07-30 15:19:39
使用Java内置类TimeUnit可以更好地处理日期差转换。它提供了实现这一点的实用方法：
Date startDate = // Set start date
Date endDate = // Set end date
long 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);Stack Overflow用户
发布于 2013-07-30 14:58:18
使用Joda-Time库
DateTime startTime, endTime;
Period p = new Period(startTime, endTime);
long hours = p.getHours();
long minutes = p.getMinutes();Joda时间有一个时间间隔的概念：
Interval interval = new Interval(oldTime, new Instant());再举一个Date Difference示例
再来一个Link
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();页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/17940200
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