当用户在我的程序中开始某些操作时，我想使用 System.currentTimeMillis() 记录时间 . 当他完成时，我将从 start 变量中减去当前的 System.currentTimeMillis() ，并且我想向他们展示使用人类可读格式（例如"XX hours, XX mins, XX seconds"或甚至"XX mins, XX seconds"）所经过的时间，因为它不可能花费一个小时的时间 .

最好的方法是什么？

25 回答

这是基于布伦特纳什答案的答案，希望有所帮助！

public static String getDurationBreakdown(long millis)
{
    String[] units = {" Days ", " Hours ", " Minutes ", " Seconds "};
    Long[] values = new Long[units.length];
    if(millis < 0)
    {
        throw new IllegalArgumentException("Duration must be greater than zero!");
    }

    values[0] = TimeUnit.MILLISECONDS.toDays(millis);
    millis -= TimeUnit.DAYS.toMillis(values[0]);
    values[1] = TimeUnit.MILLISECONDS.toHours(millis);
    millis -= TimeUnit.HOURS.toMillis(values[1]);
    values[2] = TimeUnit.MILLISECONDS.toMinutes(millis);
    millis -= TimeUnit.MINUTES.toMillis(values[2]);
    values[3] = TimeUnit.MILLISECONDS.toSeconds(millis);

    StringBuilder sb = new StringBuilder(64);
    boolean startPrinting = false;
    for(int i = 0; i < units.length; i++){
        if( !startPrinting && values[i] != 0)
            startPrinting = true;
        if(startPrinting){
            sb.append(values[i]);
            sb.append(units[i]);
        }
    }

    return(sb.toString());
}

回复于 2024-05-02T01:29:32+08:00

Joda-Time

使用Joda-Time：

DateTime startTime = new DateTime();

// do something

DateTime endTime = new DateTime();
Duration duration = new Duration(startTime, endTime);
Period period = duration.toPeriod().normalizedStandard(PeriodType.time());
System.out.println(PeriodFormat.getDefault().print(period));

回复于 2024-05-02T01:29:32+08:00

1
嗯...一秒钟多少毫秒？一会儿？分工不是那么难 .
```
int seconds = (int) ((milliseconds / 1000) % 60);
int minutes = (int) ((milliseconds / 1000) / 60);
```
像这样继续几小时，几天，几周，几个月，一年，几十年，等等 .
回复于 2024-05-02T01:29:32+08:00

重新访问@ brent-nash贡献，我们可以使用模数函数而不是减法，并使用String.format方法作为结果字符串：

/**
   * Convert a millisecond duration to a string format
   * 
   * @param millis A duration to convert to a string form
   * @return A string of the form "X Days Y Hours Z Minutes A Seconds B Milliseconds".
   */
   public static String getDurationBreakdown(long millis) {
       if (millis < 0) {
          throw new IllegalArgumentException("Duration must be greater than zero!");
       }

       long days = TimeUnit.MILLISECONDS.toDays(millis);
       long hours = TimeUnit.MILLISECONDS.toHours(millis) % 24;
       long minutes = TimeUnit.MILLISECONDS.toMinutes(millis) % 60;
       long seconds = TimeUnit.MILLISECONDS.toSeconds(millis) % 60;
       long milliseconds = millis % 1000;

       return String.format("%d Days %d Hours %d Minutes %d Seconds %d Milliseconds",
                            days, hours, minutes, seconds, milliseconds);
   }

回复于 2024-05-02T01:29:32+08:00

适用于API 9以下的Android

(String.format("%d hr %d min, %d sec", millis/(1000*60*60), (millis%(1000*60*60))/(1000*60), ((millis%(1000*60*60))%(1000*60))/1000))

回复于 2024-05-02T01:29:32+08:00

使用java.util.concurrent.TimeUnit类：

String.format("%d min, %d sec", 
    TimeUnit.MILLISECONDS.toMinutes(millis),
    TimeUnit.MILLISECONDS.toSeconds(millis) - 
    TimeUnit.MINUTES.toSeconds(TimeUnit.MILLISECONDS.toMinutes(millis))
);

注意： TimeUnit 是Java 1.5规范的一部分，但是从Java 1.6开始添加了 toMinutes .

要为值0-9添加前导零，只需执行以下操作：

String.format("%02d min, %02d sec", 
    TimeUnit.MILLISECONDS.toMinutes(millis),
    TimeUnit.MILLISECONDS.toSeconds(millis) - 
    TimeUnit.MINUTES.toSeconds(TimeUnit.MILLISECONDS.toMinutes(millis))
);

如果不支持 TimeUnit 或 toMinutes （例如在API版本9之前的Android上），请使用以下等式：

int seconds = (int) (milliseconds / 1000) % 60 ;
int minutes = (int) ((milliseconds / (1000*60)) % 60);
int hours   = (int) ((milliseconds / (1000*60*60)) % 24);
//etc...

回复于 2024-05-02T01:29:32+08:00

5

我不会仅仅为了那个拉出额外的依赖（毕竟分裂并不那么难），但是如果你还在使用Commons Lang，那么就有DurationFormatUtils .

回复于 2024-05-02T01:29:32+08:00

如果您知道时差不到一小时，那么您可以使用以下代码：

Calendar c1 = Calendar.getInstance();
    Calendar c2 = Calendar.getInstance();

    c2.add(Calendar.MINUTE, 51);

    long diff = c2.getTimeInMillis() - c1.getTimeInMillis();

    c2.set(Calendar.MINUTE, 0);
    c2.set(Calendar.HOUR, 0);
    c2.set(Calendar.SECOND, 0);

    DateFormat df = new SimpleDateFormat("mm:ss");
    long diff1 = c2.getTimeInMillis() + diff;
    System.out.println(df.format(new Date(diff1)));

结果将是：51：00

回复于 2024-05-02T01:29:32+08:00

1
只是为了添加更多信息，如果你想格式化如下：HH：mm：ss

0 <= HH <=无穷大

0 <= mm <60

0 <= ss <60

用这个：
```
int h = (int) ((startTimeInMillis / 1000) / 3600);
int m = (int) (((startTimeInMillis / 1000) / 60) % 60);
int s = (int) ((startTimeInMillis / 1000) % 60);
```
我现在才遇到这个问题，并想出了这个问题
回复于 2024-05-02T01:29:32+08:00

113

手划分，或使用SimpleDateFormat API .

long start = System.currentTimeMillis();
// do your work...
long elapsed = System.currentTimeMillis() - start;
DateFormat df = new SimpleDateFormat("HH 'hours', mm 'mins,' ss 'seconds'");
df.setTimeZone(TimeZone.getTimeZone("GMT+0"));
System.out.println(df.format(new Date(elapsed)));

Edit by Bombe ：评论中显示，这种方法仅适用于较小的持续时间（即不到一天） .

回复于 2024-05-02T01:29:32+08:00

这在Java 9中更容易：

Duration elapsedTime = Duration.ofMillis(millisDiff );
    String humanReadableElapsedTime = String.format(
            "%d hours, %d mins, %d seconds",
            elapsedTime.toHours(),
            elapsedTime.toMinutesPart(),
            elapsedTime.toSecondsPart());

这会产生类似 0 hours, 39 mins, 9 seconds 的字符串 .

如果要在格式化之前舍入到整秒：

elapsedTime = elapsedTime.plusMillis(500).truncatedTo(ChronoUnit.SECONDS);

如果它们是0，则省略小时：

long hours = elapsedTime.toHours();
    String humanReadableElapsedTime;
    if (hours == 0) {
        humanReadableElapsedTime = String.format(
                "%d mins, %d seconds",
                elapsedTime.toMinutesPart(),
                elapsedTime.toSecondsPart());

    } else {
        humanReadableElapsedTime = String.format(
                "%d hours, %d mins, %d seconds",
                hours,
                elapsedTime.toMinutesPart(),
                elapsedTime.toSecondsPart());
    }

现在我们可以举例如 39 mins, 9 seconds .

要使用前导零打印分钟和秒以使它们始终为两位数，只需将 02 插入相关的格式说明符中，即：

String humanReadableElapsedTime = String.format(
            "%d hours, %02d mins, %02d seconds",
            elapsedTime.toHours(),
            elapsedTime.toMinutesPart(),
            elapsedTime.toSecondsPart());

现在我们可以举例如 0 hours, 39 mins, 09 seconds .

回复于 2024-05-02T01:29:32+08:00

4
Shortest solution:

这可能是最短的也涉及时区 .
```
System.out.printf("%tT", millis-TimeZone.getDefault().getRawOffset());
```
以哪些输出为例：
```
00:18:32
```
Explanation:

%tT 是将24小时制格式化为 %tH:%tM:%tS 的时间 .

%tT 也接受long作为输入，因此无需创建 Date . printf() 将只打印以毫秒为单位指定的时间，但在当前时区中，我们必须减去当前时区的原始偏移量，使0毫秒为0小时，而不是当前时区的时间偏移值 .

Note #1: 如果您需要将结果作为 String ，您可以这样得到：
```
String t = String.format("%tT", millis-TimeZone.getDefault().getRawOffset());
```
Note #2: 如果 millis 小于一天，则仅提供正确的结果，因为日期部分未包含在输出中 .
回复于 2024-05-02T01:29:32+08:00

我认为最好的方法是：

String.format("%d min, %d sec", 
    TimeUnit.MILLISECONDS.toSeconds(length)/60,
    TimeUnit.MILLISECONDS.toSeconds(length) % 60 );

回复于 2024-05-02T01:29:32+08:00

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

回复于 2024-05-02T01:29:32+08:00

long time = 1536259;

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

打印：

25：36：259

回复于 2024-05-02T01:29:32+08:00

34
在Java 8中使用java.time package：
```
Instant start = Instant.now();
Thread.sleep(63553);
Instant end = Instant.now();
System.out.println(Duration.between(start, end));
```
输出位于ISO 8601 Duration format： PT1M3.553S （1分3.553秒） .
回复于 2024-05-02T01:29:32+08:00

这个答案类似于上面的一些答案 . 但是，我认为这将是有益的，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);
}

回复于 2024-05-02T01:29:32+08:00

我的简单计算：

String millisecToTime(int millisec) {
    int sec = millisec/1000;
    int second = sec % 60;
    int minute = sec / 60;
    if (minute >= 60) {
        int hour = minute / 60;
        minute %= 60;
        return hour + ":" + (minute < 10 ? "0" + minute : minute) + ":" + (second < 10 ? "0" + second : second);
    }
    return minute + ":" + (second < 10 ? "0" + second : second);
}

快乐编码:)

回复于 2024-05-02T01:29:32+08:00

1113

我已在另一个answer中介绍了这一点，但您可以这样做：

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;
}

输出类似于 Map:{DAYS=1, HOURS=3, MINUTES=46, SECONDS=40, MILLISECONDS=0, MICROSECONDS=0, NANOSECONDS=0} ，单位已订购 .

由您决定如何根据目标语言环境来国际化此数据 .

回复于 2024-05-02T01:29:32+08:00

这儿存在一个问题 . 当毫秒为59999时，实际上它是1分钟，但它将被计算为59秒并且丢失999毫秒 .

以下是基于之前答案的修改版本，可以解决此问题：

public static String formatTime(long millis) {
    long seconds = Math.round((double) millis / 1000);
    long hours = TimeUnit.SECONDS.toHours(seconds);
    if (hours > 0)
        seconds -= TimeUnit.HOURS.toSeconds(hours);
    long minutes = seconds > 0 ? TimeUnit.SECONDS.toMinutes(seconds) : 0;
    if (minutes > 0)
        seconds -= TimeUnit.MINUTES.toSeconds(minutes);
    return hours > 0 ? String.format("%02d:%02d:%02d", hours, minutes, seconds) : String.format("%02d:%02d", minutes, seconds);
}

回复于 2024-05-02T01:29:32+08:00

根据@ siddhadev的回答，我编写了一个将毫秒转换为格式化字符串的函数：

/**
     * Convert a millisecond duration to a string format
     * 
     * @param millis A duration to convert to a string form
     * @return A string of the form "X Days Y Hours Z Minutes A Seconds".
     */
    public static String getDurationBreakdown(long millis) {
        if(millis < 0) {
            throw new IllegalArgumentException("Duration must be greater than zero!");
        }

        long days = TimeUnit.MILLISECONDS.toDays(millis);
        millis -= TimeUnit.DAYS.toMillis(days);
        long hours = TimeUnit.MILLISECONDS.toHours(millis);
        millis -= TimeUnit.HOURS.toMillis(hours);
        long minutes = TimeUnit.MILLISECONDS.toMinutes(millis);
        millis -= TimeUnit.MINUTES.toMillis(minutes);
        long seconds = TimeUnit.MILLISECONDS.toSeconds(millis);

        StringBuilder sb = new StringBuilder(64);
        sb.append(days);
        sb.append(" Days ");
        sb.append(hours);
        sb.append(" Hours ");
        sb.append(minutes);
        sb.append(" Minutes ");
        sb.append(seconds);
        sb.append(" Seconds");

        return(sb.toString());
    }

回复于 2024-05-02T01:29:32+08:00

我修改了@MyKuLLSKI的答案并添加了plurlization支持 . 我花了几秒钟，因为我不需要它们，但如果你需要，可以随意重新添加它 .

public static String intervalToHumanReadableTime(int intervalMins) {

    if(intervalMins <= 0) {
        return "0";
    } else {

        long intervalMs = intervalMins * 60 * 1000;

        long days = TimeUnit.MILLISECONDS.toDays(intervalMs);
        intervalMs -= TimeUnit.DAYS.toMillis(days);
        long hours = TimeUnit.MILLISECONDS.toHours(intervalMs);
        intervalMs -= TimeUnit.HOURS.toMillis(hours);
        long minutes = TimeUnit.MILLISECONDS.toMinutes(intervalMs);

        StringBuilder sb = new StringBuilder(12);

        if (days >= 1) {
            sb.append(days).append(" day").append(pluralize(days)).append(", ");
        }

        if (hours >= 1) {
            sb.append(hours).append(" hour").append(pluralize(hours)).append(", ");
        }

        if (minutes >= 1) {
            sb.append(minutes).append(" minute").append(pluralize(minutes));
        } else {
            sb.delete(sb.length()-2, sb.length()-1);
        }

        return(sb.toString());          

    }

}

public static String pluralize(long val) {
    return (Math.round(val) > 1 ? "s" : "");
}

回复于 2024-05-02T01:29:32+08:00

对于不到一小时的小时间，我更喜欢：

long millis = ...

System.out.printf("%1$TM:%1$TS", millis);
// or
String str = String.format("%1$TM:%1$TS", millis);

更长的intervalls：

private static final long HOUR = TimeUnit.HOURS.toMillis(1);
...
if (millis < HOUR) {
    System.out.printf("%1$TM:%1$TS%n", millis);
} else {
    System.out.printf("%d:%2$TM:%2$TS%n", millis / HOUR, millis % HOUR);
}

回复于 2024-05-02T01:29:32+08:00

使用java.util.concurrent.TimeUnit，并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));

回复于 2024-05-02T01:29:32+08:00

对于正确的字符串（“1小时，3秒”，“3分钟”但不是“0小时，0分钟，3秒”）我写这段代码：

int seconds = (int)(millis / 1000) % 60 ;
int minutes = (int)((millis / (1000*60)) % 60);
int hours = (int)((millis / (1000*60*60)) % 24);
int days = (int)((millis / (1000*60*60*24)) % 365);
int years = (int)(millis / 1000*60*60*24*365);

ArrayList<String> timeArray = new ArrayList<String>();

if(years > 0)   
    timeArray.add(String.valueOf(years)   + "y");

if(days > 0)    
    timeArray.add(String.valueOf(days) + "d");

if(hours>0)   
    timeArray.add(String.valueOf(hours) + "h");

if(minutes>0) 
    timeArray.add(String.valueOf(minutes) + "min");

if(seconds>0) 
    timeArray.add(String.valueOf(seconds) + "sec");

String time = "";
for (int i = 0; i < timeArray.size(); i++) 
{
    time = time + timeArray.get(i);
    if (i != timeArray.size() - 1)
        time = time + ", ";
}

if (time == "")
  time = "0 sec";

回复于 2024-05-02T01:29:32+08:00

如何在Java中将毫秒转换为“X分钟，x秒”？

25 回答

Joda-Time

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