如果你用d1和d2作为日期，最好的解决方案可能是这样的:

int days1 = d1.getTime()/(60*60*24*1000);//find the number of days since the epoch.
int days2 = d2.getTime()/(60*60*24*1000);

然后说

days2-days1

或者其他

有很多方法可以找到日期和时间之间的区别。我所知道的最简单的方法之一是:

      Calendar calendar1 = Calendar.getInstance();
      Calendar calendar2 = Calendar.getInstance();
      calendar1.set(2012, 04, 02);
      calendar2.set(2012, 04, 04);
      long milsecs1= calendar1.getTimeInMillis();
      long milsecs2 = calendar2.getTimeInMillis();
      long diff = milsecs2 - milsecs1;
      long dsecs = diff / 1000;
      long dminutes = diff / (60 * 1000);
      long dhours = diff / (60 * 60 * 1000);
      long ddays = diff / (24 * 60 * 60 * 1000);

      System.out.println("Your Day Difference="+ddays);

打印语句只是一个示例—您可以按照自己喜欢的方式格式化它。

由于这里所有的答案都是正确的，但使用传统java或第三方库，如joda或类似的，我将放弃使用新java的另一种方式。Java 8及以后版本中的时间类。参见Oracle教程。

使用LocalDate和ChronoUnit:

LocalDate d1 = LocalDate.of(2017, 5, 1);
LocalDate d2 = LocalDate.of(2017, 5, 18);

long days = ChronoUnit.DAYS.between(d1, d2);
System.out.println( days );

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

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);

使用GMT时区获取一个Calendar实例，使用Calendar类的set方法设置时间。GMT时区偏移量为0(并不重要)，夏令时标志设置为false。

    final Calendar cal = Calendar.getInstance(TimeZone.getTimeZone("GMT"));

    cal.set(Calendar.YEAR, 2011);
    cal.set(Calendar.MONTH, 9);
    cal.set(Calendar.DAY_OF_MONTH, 29);
    cal.set(Calendar.HOUR, 0);
    cal.set(Calendar.MINUTE, 0);
    cal.set(Calendar.SECOND, 0);
    final Date startDate = cal.getTime();

    cal.set(Calendar.YEAR, 2011);
    cal.set(Calendar.MONTH, 12);
    cal.set(Calendar.DAY_OF_MONTH, 21);
    cal.set(Calendar.HOUR, 0);
    cal.set(Calendar.MINUTE, 0);
    cal.set(Calendar.SECOND, 0);
    final Date endDate = cal.getTime();

    System.out.println((endDate.getTime() - startDate.getTime()) % (1000l * 60l * 60l * 24l));