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?

由于这里所有的答案都是正确的，但使用传统java或第三方库，如joda或类似的，我将放弃使用新java的另一种方式。Java 8及以后版本中的时间类。参见Oracle教程。

使用LocalDate和ChronoUnit:

LocalDate d1 = LocalDate.of(2017, 5, 1);
LocalDate d2 = LocalDate.of(2017, 5, 18);

long days = ChronoUnit.DAYS.between(d1, d2);
System.out.println( days );

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 

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;        
   }

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
}

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