我在Scala中使用Java的Java .util.Date类,并希望比较Date对象和当前时间。我知道我可以通过使用getTime()来计算delta:

(new java.util.Date()).getTime() - oldDate.getTime()

然而,这只给我留下一个长表示毫秒。有没有更简单,更好的方法来得到时间?


当前回答

查看示例http://www.roseindia.net/java/beginners/DateDifferent.shtml 这个例子给出了天、小时、分钟、秒和毫秒的差异:)。

import java.util.Calendar;
import java.util.Date;

public class DateDifferent {
    public static void main(String[] args) {
        Date date1 = new Date(2009, 01, 10);
        Date date2 = new Date(2009, 07, 01);
        Calendar calendar1 = Calendar.getInstance();
        Calendar calendar2 = Calendar.getInstance();
        calendar1.setTime(date1);
        calendar2.setTime(date2);
        long milliseconds1 = calendar1.getTimeInMillis();
        long milliseconds2 = calendar2.getTimeInMillis();
        long diff = milliseconds2 - milliseconds1;
        long diffSeconds = diff / 1000;
        long diffMinutes = diff / (60 * 1000);
        long diffHours = diff / (60 * 60 * 1000);
        long diffDays = diff / (24 * 60 * 60 * 1000);
        System.out.println("\nThe Date Different Example");
        System.out.println("Time in milliseconds: " + diff + " milliseconds.");
        System.out.println("Time in seconds: " + diffSeconds + " seconds.");
        System.out.println("Time in minutes: " + diffMinutes + " minutes.");
        System.out.println("Time in hours: " + diffHours + " hours.");
        System.out.println("Time in days: " + diffDays + " days.");
    }
}

其他回答

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

先回答最初的问题:

将以下代码放入Long getAge(){}这样的函数中

Date dahora = new Date();
long MillisToYearsByDiv = 1000l *60l * 60l * 24l * 365l;
long javaOffsetInMillis = 1990l * MillisToYearsByDiv;
long realNowInMillis = dahora.getTime() + javaOffsetInMillis;
long realBirthDayInMillis = this.getFechaNac().getTime() + javaOffsetInMillis;
long ageInMillis = realNowInMillis - realBirthDayInMillis;

return ageInMillis / MillisToYearsByDiv;

这里最重要的是在乘法和除法时处理长数字。当然,还有Java在日期演算中应用的偏移量。

:)

这可能是最直接的方法了——也许是因为我已经用Java编写了一段时间了(它的日期和时间库确实很笨拙),但对我来说,代码看起来“简单而漂亮”!

您是否对以毫秒为单位返回的结果感到满意,或者您的问题的一部分是希望以某种替代格式返回?

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));

如果你需要一个格式化的返回字符串 “2天03h 42m 07s”,试试这个:

public String fill2(int value)
{
    String ret = String.valueOf(value);

    if (ret.length() < 2)
        ret = "0" + ret;            
    return ret;
}

public String get_duration(Date date1, Date date2)
{                   
    TimeUnit timeUnit = TimeUnit.SECONDS;

    long diffInMilli = date2.getTime() - date1.getTime();
    long s = timeUnit.convert(diffInMilli, TimeUnit.MILLISECONDS);

    long days = s / (24 * 60 * 60);
    long rest = s - (days * 24 * 60 * 60);
    long hrs = rest / (60 * 60);
    long rest1 = rest - (hrs * 60 * 60);
    long min = rest1 / 60;      
    long sec = s % 60;

    String dates = "";
    if (days > 0) dates = days + " Days ";

    dates += fill2((int) hrs) + "h ";
    dates += fill2((int) min) + "m ";
    dates += fill2((int) sec) + "s ";

    return dates;
}