如何在java中找到兩個日期之間的差異持續時間？我有兩個DateTime對象，需要找到它們的差異持續時間，我有以下代碼，但不知道如何繼續它以獲得預期的結果如下：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碼 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.");
3 回答
函數式編程
TA貢獻1807條經驗 獲得超9個贊
嘗試以下方法
{
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;}
慕尼黑8549860
TA貢獻1818條經驗 獲得超11個贊
可以使用Java內置類TimeUnit以更好的方式處理日期差異轉換。它提供了實用方法：
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);
茅侃侃
TA貢獻1842條經驗 獲得超22個贊
使用Joda-Time庫
DateTime startTime, endTime;Period p = new Period(startTime, endTime);long hours = p.getHours();long minutes = p.getMinutes();
Joda Time有一個時間間隔的概念：
Interval interval = new Interval(oldTime, new Instant());
另一個例子 日期差異
還有一個鏈接
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();
添加回答
舉報
0/150
提交
取消