另一个简单的比较方法是基于这里的答案和我的导师的指导

public static int compare(Date d1, Date d2) {
    Calendar c1 = Calendar.getInstance();
    Calendar c2 = Calendar.getInstance();
    c1.setTime(d1);
    c1.set(Calendar.MILLISECOND, 0);
    c1.set(Calendar.SECOND, 0);
    c1.set(Calendar.MINUTE, 0);
    c1.set(Calendar.HOUR_OF_DAY, 0);
    c2.setTime(d2);
    c2.set(Calendar.MILLISECOND, 0);
    c2.set(Calendar.SECOND, 0);
    c2.set(Calendar.MINUTE, 0);
    c2.set(Calendar.HOUR_OF_DAY, 0);
    return c1.getTime().compareTo(c2.getTime());
  }

编辑: 根据@Jonathan Drapeau的说法，上面的代码在某些情况下会失败(我想看看这些情况)，他建议如下:

public static int compare2(Date d1, Date d2) {
    Calendar c1 = Calendar.getInstance();
    Calendar c2 = Calendar.getInstance();
    c1.clear();
    c2.clear();
    c1.set(Calendar.YEAR, d1.getYear());
    c1.set(Calendar.MONTH, d1.getMonth());
    c1.set(Calendar.DAY_OF_MONTH, d1.getDay());
    c2.set(Calendar.YEAR, d2.getYear());
    c2.set(Calendar.MONTH, d2.getMonth());
    c2.set(Calendar.DAY_OF_MONTH, d2.getDay());
    return c1.getTime().compareTo(c2.getTime());
}

请注意，Date类已弃用，因为它不适合国际化。取而代之的是Calendar类!

我也更喜欢Joda Time，但这里有一个替代方案:

long oneDay = 24 * 60 * 60 * 1000
long d1 = first.getTime() / oneDay
long d2 = second.getTime() / oneDay
d1 == d2

EDIT

我把UTC的东西放在下面，以防你需要比较UTC以外的特定时区的日期。如果你确实有这样的需求，那么我真的建议你去找Joda。

long oneDay = 24 * 60 * 60 * 1000
long hoursFromUTC = -4 * 60 * 60 * 1000 // EST with Daylight Time Savings
long d1 = (first.getTime() + hoursFromUTC) / oneDay
long d2 = (second.getTime() + hoursFromUTC) / oneDay
d1 == d2

    Date today = new Date();
    Date endDate = new Date();//this
    endDate.setTime(endDate.getTime() - ((endDate.getHours()*60*60*1000) + (endDate.getMinutes()*60*1000) + (endDate.getSeconds()*1000)));
    today.setTime(today.getTime() - ((today.getHours()*60*60*1000) + (today.getMinutes()*60*1000) + (today.getSeconds()*1000)));

    System.out.println(endDate.compareTo(today) <= 0);

我只是将小时/分/秒设置为0，所以时间没有问题，因为现在两个日期的时间都是相同的。现在只需使用compareTo。这个方法帮助找到“if dueDate is today”，其中true表示Yes。

如果你正在寻找一个简单的解决方案，但你不想从你的项目中更改已弃用的java.util.Date类，你可以将这个方法添加到你的项目中，并继续你的探索:

使用java.util.concurrent.TimeUnit

`

public boolean isSameDay(Date first, Date second) {
    long difference_In_Time = first.getTime() - second.getTime();
        // calculate difference in days
        long difference_In_Days = 
        TimeUnit
              .MILLISECONDS
              .toDays(difference_In_Time);
        if (difference_In_Days == 0) {
            return true;
        }
        return false;
    }

`

像这样实现它:

`

Date first = ...;
Date second = ...;
if (isSameDay(first, second)) {
    // congratulations, they are the same
}
else {
   // heads up champ, they are not the same
}

`

只需结合YEAR属性检查DAY_OF_YEAR

boolean isSameDay = 
firstCal.get(Calendar.YEAR) == secondCal.get(Calendar.YEAR) &&
firstCal.get(Calendar.DAY_OF_YEAR) == secondCal.get(Calendar.DAY_OF_YEAR)

编辑:

现在我们可以使用Kotlin扩展函数的强大功能

fun Calendar.isSameDay(second: Calendar): Boolean {

    return this[Calendar.YEAR] == second[Calendar.YEAR] && this[Calendar.DAY_OF_YEAR] == second[Calendar.DAY_OF_YEAR]
}

fun Calendar.compareDatesOnly(other: Calendar): Int {

    return when {
        isSameDay(other) -> 0
        before(other) -> -1
        else -> 1
    }
}

我更倾向于直接使用Joda库而不是java.util.Date，因为Joda区分了日期和时间(参见YearMonthDay和DateTime类)。

然而，如果你确实希望使用java.util.Date，我建议写一个实用方法;如。

public static Date setTimeToMidnight(Date date) {
    Calendar calendar = Calendar.getInstance();

    calendar.setTime( date );
    calendar.set(Calendar.HOUR_OF_DAY, 0);
    calendar.set(Calendar.MINUTE, 0);
    calendar.set(Calendar.SECOND, 0);
    calendar.set(Calendar.MILLISECOND, 0);

    return calendar.getTime();
}