我在Scala中使用Java的Java .util.Date类,并希望比较Date对象和当前时间。我知道我可以通过使用getTime()来计算delta:

(new java.util.Date()).getTime() - oldDate.getTime()

然而,这只给我留下一个长表示毫秒。有没有更简单,更好的方法来得到时间?


当前回答

这可能是最直接的方法了——也许是因为我已经用Java编写了一段时间了(它的日期和时间库确实很笨拙),但对我来说,代码看起来“简单而漂亮”!

您是否对以毫秒为单位返回的结果感到满意,或者您的问题的一部分是希望以某种替代格式返回?

其他回答

@Michael Borgwardt的回答实际上在Android上不能正常运行。存在舍入错误。例如5月19日到21日是1天,因为它是1.99:1。在转换为int型之前使用round。

Fix

int diffInDays = (int)Math.round(( (newerDate.getTime() - olderDate.getTime()) 
                 / (1000 * 60 * 60 * 24) ))

请注意,这适用于UTC日期,因此如果查看本地日期,差异可能会减小一天。由于夏时制的原因,要让它在本地日期上正确工作,需要一种完全不同的方法。

让我来看看Joda Interval和Days之间的差异:

DateTime start = new DateTime(2012, 2, 6, 10, 44, 51, 0);
DateTime end = new DateTime(2012, 2, 6, 11, 39, 47, 1);
Interval interval = new Interval(start, end);
Period period = interval.toPeriod();
System.out.println(period.getYears() + " years, " + period.getMonths() + " months, " + period.getWeeks() + " weeks, " + period.getDays() + " days");
System.out.println(period.getHours() + " hours, " + period.getMinutes() + " minutes, " + period.getSeconds() + " seconds ");
//Result is:
//0 years, 0 months, *1 weeks, 1 days*
//0 hours, 54 minutes, 56 seconds 

//Period can set PeriodType,such as PeriodType.yearMonthDay(),PeriodType.yearDayTime()...
Period p = new Period(start, end, PeriodType.yearMonthDayTime());
System.out.println(p.getYears() + " years, " + p.getMonths() + " months, " + p.getWeeks() + " weeks, " + p.getDays() + "days");
System.out.println(p.getHours() + " hours, " + p.getMinutes() + " minutes, " + p.getSeconds() + " seconds ");
//Result is:
//0 years, 0 months, *0 weeks, 8 days*
//0 hours, 54 minutes, 56 seconds 
Days d = Days.daysBetween(startDate, endDate);
int days = d.getDays();

https://www.joda.org/joda-time/faq.html#datediff

以下是一种解决方案,因为我们有许多方法可以实现这一点:

  import java.util.*; 
   int syear = 2000;
   int eyear = 2000;
   int smonth = 2;//Feb
   int emonth = 3;//Mar
   int sday = 27;
   int eday = 1;
   Date startDate = new Date(syear-1900,smonth-1,sday);
   Date endDate = new Date(eyear-1900,emonth-1,eday);
   int difInDays = (int) ((endDate.getTime() - startDate.getTime())/(1000*60*60*24));

以毫秒为单位减去日期是可行的(如另一篇文章所述),但在清除日期的时间部分时,你必须使用HOUR_OF_DAY而不是HOUR:

public static final long MSPERDAY = 60 * 60 * 24 * 1000;
...
final Calendar dateStartCal = Calendar.getInstance();
dateStartCal.setTime(dateStart);
dateStartCal.set(Calendar.HOUR_OF_DAY, 0); // Crucial.
dateStartCal.set(Calendar.MINUTE, 0);
dateStartCal.set(Calendar.SECOND, 0);
dateStartCal.set(Calendar.MILLISECOND, 0);
final Calendar dateEndCal = Calendar.getInstance();
dateEndCal.setTime(dateEnd);
dateEndCal.set(Calendar.HOUR_OF_DAY, 0); // Crucial.
dateEndCal.set(Calendar.MINUTE, 0);
dateEndCal.set(Calendar.SECOND, 0);
dateEndCal.set(Calendar.MILLISECOND, 0);
final long dateDifferenceInDays = ( dateStartCal.getTimeInMillis()
                                  - dateEndCal.getTimeInMillis()
                                  ) / MSPERDAY;
if (dateDifferenceInDays > 15) {
    // Do something if difference > 15 days
}