我想有一个compareTo方法来忽略java.util.Date的时间部分。我想有很多方法可以解决这个问题。最简单的方法是什么?


当前回答

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!

其他回答

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

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

    return calendar.getTime();
}

这将帮助你在不考虑时间的情况下比较日期。

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
}

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
}

`

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。

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));