int daysDiff = (date1.getTime() - date2.getTime()) / MILLIS_PER_DAY;

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));

注意:startDate和endDates为-> java.util.Date

import org.joda.time.Duration;
import org.joda.time.Interval;
// Use .getTime() unless it is a joda DateTime object
Interval interval = new Interval(startDate.getTime(), endDate.getTime());
Duration period = interval.toDuration();
//gives the number of days elapsed between start and end date.
period.getStandardDays();

与天类似，你也可以得到小时、分钟和秒

period.getStandardHours();
period.getStandardMinutes();
period.getStandardSeconds();

在阅读了许多关于这个问题的回答和评论后，我的印象是，要么使用Joda时间，要么考虑到日光节约时间的一些特点等等。由于这两种方法我都不想做，所以我最终编写了几行代码来计算两个日期之间的差异，而没有使用任何与日期或时间相关的Java类。

在下面的代码中，年、月和日的数字与现实生活中的数字相同。例如，2015年12月24日，年= 2015，月= 12，日= 24。

我想分享这些代码，以防其他人想要使用它。有3种方法:1)找出给定年份是否是闰年的方法2)计算给定年份1月1日的天数的方法3)计算任意两个日期之间天数的方法2(结束日期减去开始日期)。

方法如下:

1)

public static boolean isLeapYear (int year) {
    //Every 4. year is a leap year, except if the year is divisible by 100 and not by 400
    //For example 1900 is not a leap year but 2000 is

    boolean result = false;

    if (year % 4 == 0) {
        result = true;
    }
    if (year % 100 == 0) {
        result = false;
    }
    if (year % 400 == 0) {
        result = true;
    }

    return result;

}

2)

public static int daysGoneSince (int yearZero, int year, int month, int day) {
    //Calculates the day number of the given date; day 1 = January 1st in the yearZero

    //Validate the input
    if (year < yearZero || month < 1 || month > 12 || day < 1 || day > 31) {
        //Throw an exception
        throw new IllegalArgumentException("Too many or too few days in month or months in year or the year is smaller than year zero");
    }
    else if (month == 4 || month == 6 || month == 9 || month == 11) {//Months with 30 days
        if (day == 31) {
            //Throw an exception
            throw new IllegalArgumentException("Too many days in month");
        }
    }
    else if (month == 2) {//February 28 or 29
        if (isLeapYear(year)) {
            if (day > 29) {
                //Throw an exception
                throw new IllegalArgumentException("Too many days in month");
            }
        }
        else if (day > 28) {
            //Throw an exception
            throw new IllegalArgumentException("Too many days in month");
        }
    }

    //Start counting days
    int days = 0;

    //Days in the target month until the target day
    days = days + day;

    //Days in the earlier months in the target year
    for (int i = 1; i < month; i++) {
        switch (i) {
            case 1: case 3: case 5:
            case 7: case 8: case 10:
            case 12:
                days = days + 31;
                break;
            case 2:
                days = days + 28;
                if (isLeapYear(year)) {
                    days = days + 1;
                }
                break;
            case 4: case 6: case 9: case 11:
                days = days + 30;
                break;
        }
    }

    //Days in the earlier years
    for (int i = yearZero; i < year; i++) {
        days = days + 365;
        if (isLeapYear(i)) {
            days = days + 1;
        }
    }

    return days;

}

3)

public static int dateDiff (int startYear, int startMonth, int startDay, int endYear, int endMonth, int endDay) {

    int yearZero;

    //daysGoneSince presupposes that the first argument be smaller or equal to the second argument
    if (10000 * startYear + 100 * startMonth + startDay > 10000 * endYear + 100 * endMonth + endDay) {//If the end date is earlier than the start date
        yearZero = endYear;
    }
    else {
        yearZero = startYear;
    }

    return daysGoneSince(yearZero, endYear, endMonth, endDay) - daysGoneSince(yearZero, startYear, startMonth, startDay);

}

因为您正在使用Scala，所以有一个非常好的Scala库Lamma。在南丫岛，你可以直接用-运算符减去日期

scala> Date(2015, 5, 5) - 2     // minus days by int
res1: io.lamma.Date = Date(2015,5,3)

scala> Date(2015, 5, 15) - Date(2015, 5, 8)   // minus two days => difference between two days
res2: Int = 7

只需在两个Date对象上使用下面的方法。如果你想传递当前日期，只需传递new date()作为第二个参数，因为它是用当前时间初始化的。

public String getDateDiffString(Date dateOne, Date dateTwo)
{
    long timeOne = dateOne.getTime();
    long timeTwo = dateTwo.getTime();
    long oneDay = 1000 * 60 * 60 * 24;
    long delta = (timeTwo - timeOne) / oneDay;

    if (delta > 0) {
        return "dateTwo is " + delta + " days after dateOne";
    }
    else {
        delta *= -1;
        return "dateTwo is " + delta + " days before dateOne";
     }
}

此外，除了从天数，如果，你也想要其他参数的差异，使用下面的片段，

int year = delta / 365;
int rest = delta % 365;
int month = rest / 30;
rest = rest % 30;
int weeks = rest / 7;
int days = rest % 7;

p.s. Code完全取自SO答案。