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);

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?

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

@Michael Borgwardt的回答实际上在Android上不能正常运行。存在舍入错误。例如5月19日到21日是1天，因为它是1.99:1。在转换为int型之前使用round。

Fix

int diffInDays = (int)Math.round(( (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。

试试这个:

int epoch = (int) (new java.text.SimpleDateFormat("MM/dd/yyyy HH:mm:ss").parse("01/01/1970  00:00:00").getTime() / 1000);

你可以在parse()方法参数中编辑字符串。

一个稍微简单一点的选择:

System.currentTimeMillis() - oldDate.getTime()

至于“更好”，你到底需要什么?将时间持续时间表示为小时数和天数等的问题是，由于日期的复杂性，它可能导致不准确和错误的期望(例如，由于夏令时，一天可能有23或25小时)。