如何在Java中找到两个日期之间的差值的持续时间？

Z时代
2024-01-10
分类：问答

我有两个DateTime对象，需要找到它们之间的差值的持续时间，

我有以下代码，但不确定如何继续执行以获得预期的结果，如下所示：

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 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.");

回答：

尝试以下

{
        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;
}