如果startDate和endDate是java.util.Date的实例

我们可以从ChronoUnit enum中使用between()方法:

public long between(Temporal temporal1Inclusive, Temporal temporal2Exclusive) {
    //..
}

ChronoUnit。DAYS表示完成24小时的天数。

import java.time.temporal.ChronoUnit;

ChronoUnit.DAYS.between(startDate.toInstant(), endDate.toInstant());

//OR 

ChronoUnit.DAYS.between(Instant.ofEpochMilli(startDate.getTime()), Instant.ofEpochMilli(endDate.getTime()));

你可以使用until:

LocalDate independenceDay = LocalDate.of(2014, Month.JULY, 4);
LocalDate christmas = LocalDate.of(2014, Month.DECEMBER, 25);

System.out.println("Until christmas: " + independenceDay.until(christmas));
System.out.println("Until christmas (with crono): " + independenceDay.until(christmas, ChronoUnit.DAYS));

输出:

Until christmas: P5M21D
Until christmas (with crono): 174

如注释中所述，如果在返回Period之前没有指定单位。

文档片段:

ISO-8601日历系统中基于日期的时间量，例如“2年3个月4天”。 该类以年、月和日为单位对时间的数量或数量进行建模。有关该类的基于时间的等价物，请参见Duration。

根据VGR的评论，以下是你可以使用的方法:

ChronoUnit.DAYS.between(firstDate, secondDate)

从现在开始计算圣诞节前的天数，试试这个

System.out.println(ChronoUnit.DAYS.between(LocalDate.now(),LocalDate.of(Year.now().getValue(), Month.DECEMBER, 25)));

给你:

public class DemoDate {
    public static void main(String[] args) {
        LocalDate today = LocalDate.now();
        System.out.println("Current date: " + today);

        //add 1 month to the current date
        LocalDate date2 = today.plus(1, ChronoUnit.MONTHS);
        System.out.println("Next month: " + date2);

        // Put latest date 1st and old date 2nd in 'between' method to get -ve date difference
        long daysNegative = ChronoUnit.DAYS.between(date2, today);
        System.out.println("Days : "+daysNegative);

        // Put old date 1st and new date 2nd in 'between' method to get +ve date difference
        long datePositive = ChronoUnit.DAYS.between(today, date2);
        System.out.println("Days : "+datePositive);
    }
}

如果目标只是得到天的差异，而且既然上面的答案提到了关于委托方法，想指出的是，一旦也可以简单地使用-

public long daysInBetween(java.time.LocalDate startDate, java.time.LocalDate endDate) {
  // Check for null values here

  return endDate.toEpochDay() - startDate.toEpochDay();
}