如果你只想比较两个日期的月、日和年，下面的代码适用于我:

SimpleDateFormat sdf = new SimpleDateFormat("yyyyMMdd");
sdf.format(date1).equals(sdf.format(date2));

谢谢抢劫。

Apache commons-lang几乎无处不在。那么这个呢?

if (DateUtils.isSameDay(date1, date2)) {
    // it's same
} else if (date1.before(date2)) {
   // it's before
} else {
   // it's after
}

这里有一个来自这个博客的解决方案:http://brigitzblog.blogspot.com/2011/10/java-compare-dates.html

long milliseconds1 = calendar1.getTimeInMillis();
long milliseconds2 = calendar2.getTimeInMillis();
long diff = milliseconds2 - milliseconds1;
long diffDays = diff / (24 * 60 * 60 * 1000);
System.out.println("Time in days: " + diffDays  + " days.");

也就是说，你可以看到以毫秒为单位的时间差是否小于一天的长度。

我通过比较时间戳来解决这个问题:

    Calendar last = Calendar.getInstance();

    last.setTimeInMillis(firstTimeInMillis);

    Calendar current = Calendar.getInstance();

    if (last.get(Calendar.DAY_OF_MONTH) != current.get(Calendar.DAY_OF_MONTH)) {
        //not the same day
    }

我避免使用Joda Time，因为在Android上占用很大的空间。规模很重要。,)

对这个替代方案有什么意见吗?

SimpleDateFormat sdf = new SimpleDateFormat("yyyyMMdd");
sdf.format(date1).equals(sdf.format(date2));

使用Java 8和Instant的另一个解决方案是使用truncatedTo方法

返回截断为指定单位的此即时的副本。

例子:

@Test
public void dateTruncate() throws InterruptedException {
    Instant now = Instant.now();
    Thread.sleep(1000*5);
    Instant later = Instant.now();
    assertThat(now, not(equalTo(later)));
    assertThat(now.truncatedTo(ChronoUnit.DAYS), equalTo(later.truncatedTo(ChronoUnit.DAYS)));
}