Since dates can contain hours and minutes, final result will be rounded down, which will result in incorrect value. For example, you calculate difference between today at 22:00 p.m and day after tomorrow 00:00 a.m, so the final result will be 1, because in reality it was 1.08 or smth difference, then it gets rounded down when calling TimeUnit.MILLISECONDS.toDays(..). That's why you need to take that in account, so in my solution I subtract the remainder of milliseconds from milliseconds in a day. Additionally, if you want to count the end date, you need to +1 it.

import java.util.Date;
import java.util.concurrent.TimeUnit;

public static long getDaysBetween(Date date1, Date date2, boolean includeEndDate) {
        long millisInDay = 60 * 60 * 24 * 1000;
        long difference = Math.abs(date1.getTime() - date2.getTime());
        long add = millisInDay - (difference % millisInDay);//is used to calculate true number of days, because by default hours, minutes are also counted

        return TimeUnit.MILLISECONDS.toDays(difference + add) + (includeEndDate ? 1 : 0);
    }

测试:

Date date1 = new Date(121, Calendar.NOVEMBER, 27); //2021 Nov 27
Date date2 = new Date(121, Calendar.DECEMBER, 29); //2021 Dec 29
System.out.println( getDaysBetween(date1, date2, false) ); //32 days difference
System.out.println( getDaysBetween(date1, date2, true) ); //33 days difference

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

如果你需要一个格式化的返回字符串 “2天03h 42m 07s”，试试这个:

public String fill2(int value)
{
    String ret = String.valueOf(value);

    if (ret.length() < 2)
        ret = "0" + ret;            
    return ret;
}

public String get_duration(Date date1, Date date2)
{                   
    TimeUnit timeUnit = TimeUnit.SECONDS;

    long diffInMilli = date2.getTime() - date1.getTime();
    long s = timeUnit.convert(diffInMilli, TimeUnit.MILLISECONDS);

    long days = s / (24 * 60 * 60);
    long rest = s - (days * 24 * 60 * 60);
    long hrs = rest / (60 * 60);
    long rest1 = rest - (hrs * 60 * 60);
    long min = rest1 / 60;      
    long sec = s % 60;

    String dates = "";
    if (days > 0) dates = days + " Days ";

    dates += fill2((int) hrs) + "h ";
    dates += fill2((int) min) + "m ";
    dates += fill2((int) sec) + "s ";

    return dates;
}

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();

如果你不想使用JodaTime或类似的，最好的解决方案可能是:

final static long MILLIS_PER_DAY = 24 * 3600 * 1000;
long msDiff= date1.getTime() - date2.getTime();
long daysDiff = Math.round(msDiff / ((double)MILLIS_PER_DAY));

每天的毫秒数并不总是相同的(因为日光节约时间和闰秒)，但它非常接近，至少由于日光节约时间的偏差在较长时间内抵消了。因此，除法和舍入将给出正确的结果(至少只要所使用的本地日历不包含DST和闰秒以外的奇怪时间跳转)。

请注意，这仍然假设date1和date2被设置为一天中的同一时间。对于一天中的不同时间，你首先必须定义“日期差异”的含义，正如乔恩·斯基特指出的那样。

在涉猎了所有其他答案之后，为了保持Java 7 Date类型，但使用Java 8 diff方法更精确/标准，

public static long daysBetweenDates(Date d1, Date d2) {
    Instant instant1 = d1.toInstant();
    Instant instant2 = d2.toInstant();
    long diff = ChronoUnit.DAYS.between(instant1, instant2);
    return diff;
}