int diffInDays = (int)( (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。

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

查看示例http://www.roseindia.net/java/beginners/DateDifferent.shtml 这个例子给出了天、小时、分钟、秒和毫秒的差异:)。

import java.util.Calendar;
import java.util.Date;

public class DateDifferent {
    public static void main(String[] args) {
        Date date1 = new Date(2009, 01, 10);
        Date date2 = new Date(2009, 07, 01);
        Calendar calendar1 = Calendar.getInstance();
        Calendar calendar2 = Calendar.getInstance();
        calendar1.setTime(date1);
        calendar2.setTime(date2);
        long milliseconds1 = calendar1.getTimeInMillis();
        long milliseconds2 = calendar2.getTimeInMillis();
        long diff = milliseconds2 - milliseconds1;
        long diffSeconds = diff / 1000;
        long diffMinutes = diff / (60 * 1000);
        long diffHours = diff / (60 * 60 * 1000);
        long diffDays = diff / (24 * 60 * 60 * 1000);
        System.out.println("\nThe Date Different Example");
        System.out.println("Time in milliseconds: " + diff + " milliseconds.");
        System.out.println("Time in seconds: " + diffSeconds + " seconds.");
        System.out.println("Time in minutes: " + diffMinutes + " minutes.");
        System.out.println("Time in hours: " + diffHours + " hours.");
        System.out.println("Time in days: " + diffDays + " days.");
    }
}

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。

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);

}

不使用标准API，不行。你可以这样做:

class Duration {
    private final TimeUnit unit;
    private final long length;
    // ...
}

或者你可以使用Joda:

DateTime a = ..., b = ...;
Duration d = new Duration(a, b);