我想用H:MM:SS这样的模式以秒为单位格式化持续时间。java中当前的实用程序设计用于格式化时间,而不是持续时间。


当前回答

这是一个可行的选择。

public static String showDuration(LocalTime otherTime){          
    DateTimeFormatter df = DateTimeFormatter.ISO_LOCAL_TIME;
    LocalTime now = LocalTime.now();
    System.out.println("now: " + now);
    System.out.println("otherTime: " + otherTime);
    System.out.println("otherTime: " + otherTime.format(df));

    Duration span = Duration.between(otherTime, now);
    LocalTime fTime = LocalTime.ofNanoOfDay(span.toNanos());
    String output = fTime.format(df);

    System.out.println(output);
    return output;
}

使用

System.out.println(showDuration(LocalTime.of(9, 30, 0, 0)));

产生如下内容:

otherTime: 09:30
otherTime: 09:30:00
11:31:27.463
11:31:27.463

其他回答

我的库Time4J提供了一个基于模式的解决方案(类似于Apache DurationFormatUtils,但更灵活):

Duration<ClockUnit> duration =
    Duration.of(-573421, ClockUnit.SECONDS) // input in seconds only
    .with(Duration.STD_CLOCK_PERIOD); // performs normalization to h:mm:ss-structure
String fs = Duration.formatter(ClockUnit.class, "+##h:mm:ss").format(duration);
System.out.println(fs); // output => -159:17:01

这段代码演示了处理小时溢出和符号处理的功能,请参见基于模式的持续时间格式化程序的API。

那么下面的函数呢 + H: MM: SS 或 + H: MM: SS.sss

public static String formatInterval(final long interval, boolean millisecs )
{
    final long hr = TimeUnit.MILLISECONDS.toHours(interval);
    final long min = TimeUnit.MILLISECONDS.toMinutes(interval) %60;
    final long sec = TimeUnit.MILLISECONDS.toSeconds(interval) %60;
    final long ms = TimeUnit.MILLISECONDS.toMillis(interval) %1000;
    if( millisecs ) {
        return String.format("%02d:%02d:%02d.%03d", hr, min, sec, ms);
    } else {
        return String.format("%02d:%02d:%02d", hr, min, sec );
    }
}

在scala中,不需要库:

def prettyDuration(str:List[String],seconds:Long):List[String]={
  seconds match {
    case t if t < 60 => str:::List(s"${t} seconds")
    case t if (t >= 60 && t< 3600 ) => List(s"${t / 60} minutes"):::prettyDuration(str, t%60)
    case t if (t >= 3600 && t< 3600*24 ) => List(s"${t / 3600} hours"):::prettyDuration(str, t%3600)
    case t if (t>= 3600*24 ) => List(s"${t / (3600*24)} days"):::prettyDuration(str, t%(3600*24))
  }
}
val dur = prettyDuration(List.empty[String], 12345).mkString("")

在java8中还有另一种方法。但如果持续时间不超过24小时,则有效

public String formatDuration(Duration duration) {
    DateTimeFormatter formatter = DateTimeFormatter.ofPattern("h:mm.SSS");
    return LocalTime.ofNanoOfDay(duration.toNanos()).format(formatter);
}

使用这个func

private static String strDuration(long duration) {
    int ms, s, m, h, d;
    double dec;
    double time = duration * 1.0;

    time = (time / 1000.0);
    dec = time % 1;
    time = time - dec;
    ms = (int)(dec * 1000);

    time = (time / 60.0);
    dec = time % 1;
    time = time - dec;
    s = (int)(dec * 60);

    time = (time / 60.0);
    dec = time % 1;
    time = time - dec;
    m = (int)(dec * 60);

    time = (time / 24.0);
    dec = time % 1;
    time = time - dec;
    h = (int)(dec * 24);
    
    d = (int)time;
    
    return (String.format("%d d - %02d:%02d:%02d.%03d", d, h, m, s, ms));
}