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

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);

}

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

不幸的是，JDK Date API被严重破坏了。我推荐使用Joda Time库。

Joda Time有一个时间间隔的概念:

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

编辑:顺便说一下，Joda有两个概念:Interval表示两个时间瞬间之间的时间间隔(表示上午8点到上午10点之间的时间)，Duration表示没有实际时间边界的时间长度(例如表示两个小时!)

如果你只关心时间比较，大多数Date实现(包括JDK)实现了Comparable接口，允许你使用Comparable. compareto ()

有很多方法可以找到日期和时间之间的区别。我所知道的最简单的方法之一是:

      Calendar calendar1 = Calendar.getInstance();
      Calendar calendar2 = Calendar.getInstance();
      calendar1.set(2012, 04, 02);
      calendar2.set(2012, 04, 04);
      long milsecs1= calendar1.getTimeInMillis();
      long milsecs2 = calendar2.getTimeInMillis();
      long diff = milsecs2 - milsecs1;
      long dsecs = diff / 1000;
      long dminutes = diff / (60 * 1000);
      long dhours = diff / (60 * 60 * 1000);
      long ddays = diff / (24 * 60 * 60 * 1000);

      System.out.println("Your Day Difference="+ddays);

打印语句只是一个示例—您可以按照自己喜欢的方式格式化它。

使用java。Java 8+内置的时间框架:

ZonedDateTime now = ZonedDateTime.now();
ZonedDateTime oldDate = now.minusDays(1).minusMinutes(10);
Duration duration = Duration.between(oldDate, now);
System.out.println("ISO-8601: " + duration);
System.out.println("Minutes: " + duration.toMinutes());

输出:

ISO-8601: PT24H10M 分钟:罢工,

有关更多信息，请参阅Oracle教程和ISO 8601标准。