这个答案与上面的一些答案相似。然而，我觉得这将是有益的，因为与其他答案不同，这将删除任何额外的逗号或空格，并处理缩写。

/**
 * Converts milliseconds to "x days, x hours, x mins, x secs"
 * 
 * @param millis
 *            The milliseconds
 * @param longFormat
 *            {@code true} to use "seconds" and "minutes" instead of "secs" and "mins"
 * @return A string representing how long in days/hours/minutes/seconds millis is.
 */
public static String millisToString(long millis, boolean longFormat) {
    if (millis < 1000) {
        return String.format("0 %s", longFormat ? "seconds" : "secs");
    }
    String[] units = {
            "day", "hour", longFormat ? "minute" : "min", longFormat ? "second" : "sec"
    };
    long[] times = new long[4];
    times[0] = TimeUnit.DAYS.convert(millis, TimeUnit.MILLISECONDS);
    millis -= TimeUnit.MILLISECONDS.convert(times[0], TimeUnit.DAYS);
    times[1] = TimeUnit.HOURS.convert(millis, TimeUnit.MILLISECONDS);
    millis -= TimeUnit.MILLISECONDS.convert(times[1], TimeUnit.HOURS);
    times[2] = TimeUnit.MINUTES.convert(millis, TimeUnit.MILLISECONDS);
    millis -= TimeUnit.MILLISECONDS.convert(times[2], TimeUnit.MINUTES);
    times[3] = TimeUnit.SECONDS.convert(millis, TimeUnit.MILLISECONDS);
    StringBuilder s = new StringBuilder();
    for (int i = 0; i < 4; i++) {
        if (times[i] > 0) {
            s.append(String.format("%d %s%s, ", times[i], units[i], times[i] == 1 ? "" : "s"));
        }
    }
    return s.toString().substring(0, s.length() - 2);
}

/**
 * Converts milliseconds to "x days, x hours, x mins, x secs"
 * 
 * @param millis
 *            The milliseconds
 * @return A string representing how long in days/hours/mins/secs millis is.
 */
public static String millisToString(long millis) {
    return millisToString(millis, false);
}

long time = 1536259;

return (new SimpleDateFormat("mm:ss:SSS")).format(new Date(time));

打印:

25:36:259

我不会为此引入额外的依赖项(毕竟，除法并不那么难)，但如果您无论如何都在使用Commons Lang，还有DurationFormatUtils。

用法示例(从这里改编):

import org.apache.commons.lang3.time.DurationFormatUtils

public String getAge(long value) {
    long currentTime = System.currentTimeMillis();
    long age = currentTime - value;
    String ageString = DurationFormatUtils.formatDuration(age, "d") + "d";
    if ("0d".equals(ageString)) {
        ageString = DurationFormatUtils.formatDuration(age, "H") + "h";
        if ("0h".equals(ageString)) {
            ageString = DurationFormatUtils.formatDuration(age, "m") + "m";
            if ("0m".equals(ageString)) {
                ageString = DurationFormatUtils.formatDuration(age, "s") + "s";
                if ("0s".equals(ageString)) {
                    ageString = age + "ms";
                }
            }
        }
    }
    return ageString;
}   

例子:

long lastTime = System.currentTimeMillis() - 2000;
System.out.println("Elapsed time: " + getAge(lastTime)); 

//Output: 2s

注意:从两个LocalDateTime对象中获取millis可以使用:

long age = ChronoUnit.MILLIS.between(initTime, LocalDateTime.now())

使用java . util . concurrent。TimeUnit，使用这个简单的方法:

private static long timeDiff(Date date, Date date2, TimeUnit unit) {
    long milliDiff=date2.getTime()-date.getTime();
    long unitDiff = unit.convert(milliDiff, TimeUnit.MILLISECONDS);
    return unitDiff; 
}

例如:

SimpleDateFormat sdf = new SimpleDateFormat("yy/MM/dd HH:mm:ss");  
Date firstDate = sdf.parse("06/24/2017 04:30:00");
Date secondDate = sdf.parse("07/24/2017 05:00:15");
Date thirdDate = sdf.parse("06/24/2017 06:00:15");

System.out.println("days difference: "+timeDiff(firstDate,secondDate,TimeUnit.DAYS));
System.out.println("hours difference: "+timeDiff(firstDate,thirdDate,TimeUnit.HOURS));
System.out.println("minutes difference: "+timeDiff(firstDate,thirdDate,TimeUnit.MINUTES));
System.out.println("seconds difference: "+timeDiff(firstDate,thirdDate,TimeUnit.SECONDS));

    long startTime = System.currentTimeMillis();
    // do your work...
    long endTime=System.currentTimeMillis();
    long diff=endTime-startTime;       
    long hours=TimeUnit.MILLISECONDS.toHours(diff);
    diff=diff-(hours*60*60*1000);
    long min=TimeUnit.MILLISECONDS.toMinutes(diff);
    diff=diff-(min*60*1000);
    long seconds=TimeUnit.MILLISECONDS.toSeconds(diff);
    //hour, min and seconds variables contains the time elapsed on your work

我在另一个答案中提到了这一点，但你可以这样做:

public static Map<TimeUnit,Long> computeDiff(Date date1, Date date2) {
    long diffInMillies = date2.getTime() - date1.getTime();
    List<TimeUnit> units = new ArrayList<TimeUnit>(EnumSet.allOf(TimeUnit.class));
    Collections.reverse(units);
    Map<TimeUnit,Long> result = new LinkedHashMap<TimeUnit,Long>();
    long milliesRest = diffInMillies;
    for ( TimeUnit unit : units ) {
        long diff = unit.convert(milliesRest,TimeUnit.MILLISECONDS);
        long diffInMilliesForUnit = unit.toMillis(diff);
        milliesRest = milliesRest - diffInMilliesForUnit;
        result.put(unit,diff);
    }
    return result;
}

输出类似Map:{DAYS=1, HOURS=3, MINUTES=46, SECONDS=40, MILLISECONDS=0, MICROSECONDS=0, NANOSECONDS=0}，单位是有序的。

由您决定如何根据目标语言环境对这些数据进行国际化。