只需要对每个函数调用getTime，取其差值，然后除以一天中的毫秒数。

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;
}

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

下面的代码可以给你想要的输出:

String startDate = "Jan 01 2015";
DateTimeFormatter formatter = DateTimeFormatter.ofPattern("MMM dd yyyy");
LocalDate date = LocalDate.parse(startDate, formatter);

String currentDate = "Feb 11 2015";
LocalDate date1 = LocalDate.parse(currentDate, formatter);

System.out.println(date1.toEpochDay() - date.toEpochDay());

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);

}

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