看看Joda Time，它是一个改进的Java日期/时间API，应该可以很好地与Scala一起工作。

简单的diff(不含lib)

/**
 * Get a diff between two dates
 * @param date1 the oldest date
 * @param date2 the newest date
 * @param timeUnit the unit in which you want the diff
 * @return the diff value, in the provided unit
 */
public static long getDateDiff(Date date1, Date date2, TimeUnit timeUnit) {
    long diffInMillies = date2.getTime() - date1.getTime();
    return timeUnit.convert(diffInMillies,TimeUnit.MILLISECONDS);
}

然后你可以调用:

getDateDiff(date1,date2,TimeUnit.MINUTES);

以分钟为单位获取两个日期的差值。

TimeUnit是java.util.concurrent。TimeUnit，一个从纳米到天的标准Java枚举。

人类可读的差异(不含lib)

public static Map<TimeUnit,Long> computeDiff(Date date1, Date date2) {

    long diffInMillies = date2.getTime() - date1.getTime();

    //create the list
    List<TimeUnit> units = new ArrayList<TimeUnit>(EnumSet.allOf(TimeUnit.class));
    Collections.reverse(units);

    //create the result map of TimeUnit and difference
    Map<TimeUnit,Long> result = new LinkedHashMap<TimeUnit,Long>();
    long milliesRest = diffInMillies;

    for ( TimeUnit unit : units ) {
        
        //calculate difference in millisecond 
        long diff = unit.convert(milliesRest,TimeUnit.MILLISECONDS);
        long diffInMilliesForUnit = unit.toMillis(diff);
        milliesRest = milliesRest - diffInMilliesForUnit;

        //put the result in the map
        result.put(unit,diff);
    }

    return result;
}

http://ideone.com/5dXeu6

输出类似Map:{DAYS=1, HOURS=3, MINUTES=46, SECONDS=40, MILLISECONDS=0, MICROSECONDS=0, NANOSECONDS=0}，单位是有序的。

您只需将该映射转换为用户友好的字符串。

警告

上面的代码段计算两个瞬间之间的简单差。它会在夏令时切换期间导致问题，就像这篇文章中解释的那样。这意味着如果你计算没有时间的日期之间的差异，你可能会少了一天/小时。

在我看来，日期的差异是主观的，尤其是在日子上。你可以:

计算经过24小时的时间:day+1 - day = 1 day = 24h 计算经过的时间，考虑到夏令时:天+1 -天= 1 = 24小时(但使用午夜时间和夏令时，它可以是0天和23小时) 计算日开关的数量，这意味着一天+1 1pm -一天11am = 1天，即使经过的时间只有2h(如果有夏令时则为1h:p)

我的答案是有效的，如果你的日期差异的定义天匹配第一种情况

与JodaTime

如果你正在使用JodaTime，你可以得到2个瞬间的差异(millies支持ReadableInstant)日期:

Interval interval = new Interval(oldInstant, new Instant());

但是你也可以得到本地日期/时间的差异:

// returns 4 because of the leap year of 366 days
new Period(LocalDate.now(), LocalDate.now().plusDays(365*5), PeriodType.years()).getYears() 

// this time it returns 5
new Period(LocalDate.now(), LocalDate.now().plusDays(365*5+1), PeriodType.years()).getYears() 

// And you can also use these static methods
Years.yearsBetween(LocalDate.now(), LocalDate.now().plusDays(365*5)).getYears()

在某些地区使用毫秒方法可能会导致问题。

举个例子，03/24/2007和03/25/2007之间的差应该是1天;

然而，如果使用毫秒路径，你将得到0天，如果你在英国运行这个!

/** Manual Method - YIELDS INCORRECT RESULTS - DO NOT USE**/  
/* This method is used to find the no of days between the given dates */  
public long calculateDays(Date dateEarly, Date dateLater) {  
   return (dateLater.getTime() - dateEarly.getTime()) / (24 * 60 * 60 * 1000);  
} 

更好的实现方法是使用java.util.Calendar

/** Using Calendar - THE CORRECT WAY**/  
public static long daysBetween(Calendar startDate, Calendar endDate) {  
  Calendar date = (Calendar) startDate.clone();  
  long daysBetween = 0;  
  while (date.before(endDate)) {  
    date.add(Calendar.DAY_OF_MONTH, 1);  
    daysBetween++;  
  }  
  return daysBetween;  
}  

这是另一个样本。基本上适用于用户定义的模式。

   public static LinkedHashMap<String, Object> checkDateDiff(DateTimeFormatter dtfObj, String startDate, String endDate)
   {
          Map<String, Object> dateDiffMap = new HashMap<String, Object>();
          DateTime start = DateTime.parse(startDate,dtfObj);
          DateTime end = DateTime.parse(endDate,dtfObj);
          Interval interval = new Interval(start, end);
          Period period = interval.toPeriod();

          dateDiffMap.put("ISO-8601_PERIOD_FORMAT", period);
          dateDiffMap.put("YEAR", period.getYears());
          dateDiffMap.put("MONTH", period.getMonths());
          dateDiffMap.put("WEEK", period.getWeeks());
          dateDiffMap.put("DAY", period.getWeeks());             
          dateDiffMap.put("HOUR", period.getHours());
          dateDiffMap.put("MINUTE", period.getMinutes());
          dateDiffMap.put("SECOND", period.getSeconds());

          return dateDiffMap;        
   }

public static void main(String[] args) {

    String dateStart = "01/14/2012 09:29:58";
    String dateStop = "01/14/2012 10:31:48";

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

    Date d1 = null;
    Date d2 = null;

    try {
        d1 = format.parse(dateStart);
        d2 = format.parse(dateStop);

        DateTime date11 = new DateTime(d1);
        DateTime date22 = new DateTime(d2);
        int days = Days.daysBetween(date11.withTimeAtStartOfDay(), date22.withTimeAtStartOfDay()).getDays();
        int hours = Hours.hoursBetween(date11, date22).getHours() % 24;
        int minutes = Minutes.minutesBetween(date11, date22).getMinutes() % 60;
        int seconds = Seconds.secondsBetween(date11, date22).getSeconds() % 60;
        if (hours > 0 || minutes > 0 || seconds > 0) {
            days = days + 1;
        }

        System.out.println(days);

    } catch (Exception e) {
        e.printStackTrace();
    }

}

这将为同一天提供日期差异

如果你用d1和d2作为日期，最好的解决方案可能是这样的:

int days1 = d1.getTime()/(60*60*24*1000);//find the number of days since the epoch.
int days2 = d2.getTime()/(60*60*24*1000);

然后说

days2-days1

或者其他