这是一个正确的Java 7解决方案，没有任何依赖。

public static int countDaysBetween(Date date1, Date date2) {

    Calendar c1 = removeTime(from(date1));
    Calendar c2 = removeTime(from(date2));

    if (c1.get(YEAR) == c2.get(YEAR)) {

        return Math.abs(c1.get(DAY_OF_YEAR) - c2.get(DAY_OF_YEAR)) + 1;
    }
    // ensure c1 <= c2
    if (c1.get(YEAR) > c2.get(YEAR)) {
        Calendar c = c1;
        c1 = c2;
        c2 = c;
    }
    int y1 = c1.get(YEAR);
    int y2 = c2.get(YEAR);
    int d1 = c1.get(DAY_OF_YEAR);
    int d2 = c2.get(DAY_OF_YEAR);

    return d2 + ((y2 - y1) * 365) - d1 + countLeapYearsBetween(y1, y2) + 1;
}

private static int countLeapYearsBetween(int y1, int y2) {

    if (y1 < 1 || y2 < 1) {
        throw new IllegalArgumentException("Year must be > 0.");
    }
    // ensure y1 <= y2
    if (y1 > y2) {
        int i = y1;
        y1 = y2;
        y2 = i;
    }

    int diff = 0;

    int firstDivisibleBy4 = y1;
    if (firstDivisibleBy4 % 4 != 0) {
        firstDivisibleBy4 += 4 - (y1 % 4);
    }
    diff = y2 - firstDivisibleBy4 - 1;
    int divisibleBy4 = diff < 0 ? 0 : diff / 4 + 1;

    int firstDivisibleBy100 = y1;
    if (firstDivisibleBy100 % 100 != 0) {
        firstDivisibleBy100 += 100 - (firstDivisibleBy100 % 100);
    }
    diff = y2 - firstDivisibleBy100 - 1;
    int divisibleBy100 = diff < 0 ? 0 : diff / 100 + 1;

    int firstDivisibleBy400 = y1;
    if (firstDivisibleBy400 % 400 != 0) {
        firstDivisibleBy400 += 400 - (y1 % 400);
    }
    diff = y2 - firstDivisibleBy400 - 1;
    int divisibleBy400 = diff < 0 ? 0 : diff / 400 + 1;

    return divisibleBy4 - divisibleBy100 + divisibleBy400;
}


public static Calendar from(Date date) {

    Calendar c = Calendar.getInstance();
    c.setTime(date);

    return c;
}


public static Calendar removeTime(Calendar c) {

    c.set(HOUR_OF_DAY, 0);
    c.set(MINUTE, 0);
    c.set(SECOND, 0);
    c.set(MILLISECOND, 0);

    return c;
}

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

这可能是最直接的方法了——也许是因为我已经用Java编写了一段时间了(它的日期和时间库确实很笨拙)，但对我来说，代码看起来“简单而漂亮”!

您是否对以毫秒为单位返回的结果感到满意，或者您的问题的一部分是希望以某种替代格式返回?

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 ()

另一个纯Java变体:

public boolean isWithin30Days(Calendar queryCalendar) {

    // 1. Take the date you are checking, and roll it back N days
    Calendar queryCalMinus30Days = Calendar.getInstance();
    queryCalMinus30Days.setTime(queryCalendar.getTime());
    queryCalMinus30Days.add(Calendar.DATE, -30); // subtract 30 days from the calendar

    // 2. Get respective milliseconds for the two Calendars: now & queryCal minus N days 
    long nowL = Calendar.getInstance().getTimeInMillis();
    long queryCalMinus30DaysL = queryCalMinus30Days.getTimeInMillis();

    // 3. if nowL is still less than the queryCalMinus30DaysL, it means queryCalendar is more than 30 days into future
    boolean isWithin30Days = nowL >= queryCalMinus30DaysL;

    return isWithin30Days;
}

感谢这里的入门代码:https://stackoverflow.com/a/30207726/2162226

以毫秒为单位减去日期是可行的(如另一篇文章所述)，但在清除日期的时间部分时，你必须使用HOUR_OF_DAY而不是HOUR:

public static final long MSPERDAY = 60 * 60 * 24 * 1000;
...
final Calendar dateStartCal = Calendar.getInstance();
dateStartCal.setTime(dateStart);
dateStartCal.set(Calendar.HOUR_OF_DAY, 0); // Crucial.
dateStartCal.set(Calendar.MINUTE, 0);
dateStartCal.set(Calendar.SECOND, 0);
dateStartCal.set(Calendar.MILLISECOND, 0);
final Calendar dateEndCal = Calendar.getInstance();
dateEndCal.setTime(dateEnd);
dateEndCal.set(Calendar.HOUR_OF_DAY, 0); // Crucial.
dateEndCal.set(Calendar.MINUTE, 0);
dateEndCal.set(Calendar.SECOND, 0);
dateEndCal.set(Calendar.MILLISECOND, 0);
final long dateDifferenceInDays = ( dateStartCal.getTimeInMillis()
                                  - dateEndCal.getTimeInMillis()
                                  ) / MSPERDAY;
if (dateDifferenceInDays > 15) {
    // Do something if difference > 15 days
}