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

String duration(Temporal from, Temporal to) {
    final StringBuilder builder = new StringBuilder();
    for (ChronoUnit unit : new ChronoUnit[]{YEARS, MONTHS, WEEKS, DAYS, HOURS, MINUTES, SECONDS}) {
        long amount = unit.between(from, to);
        if (amount == 0) {
            continue;
        }
        builder.append(' ')
                .append(amount)
                .append(' ')
                .append(unit.name().toLowerCase());
        from = from.plus(amount, unit);
    }
    return builder.toString().trim();
}

其他回答

这可能有点俗气,但如果你决心使用Java 8的Java .time来实现这一点,这是一个很好的解决方案:

import java.time.Duration;
import java.time.LocalDateTime;
import java.time.format.DateTimeFormatter;
import java.time.format.DateTimeFormatterBuilder;
import java.time.temporal.ChronoField;
import java.time.temporal.Temporal;
import java.time.temporal.TemporalAccessor;
import java.time.temporal.TemporalField;
import java.time.temporal.UnsupportedTemporalTypeException;

public class TemporalDuration implements TemporalAccessor {
    private static final Temporal BASE_TEMPORAL = LocalDateTime.of(0, 1, 1, 0, 0);

    private final Duration duration;
    private final Temporal temporal;

    public TemporalDuration(Duration duration) {
        this.duration = duration;
        this.temporal = duration.addTo(BASE_TEMPORAL);
    }

    @Override
    public boolean isSupported(TemporalField field) {
        if(!temporal.isSupported(field)) return false;
        long value = temporal.getLong(field)-BASE_TEMPORAL.getLong(field);
        return value!=0L;
    }

    @Override
    public long getLong(TemporalField field) {
        if(!isSupported(field)) throw new UnsupportedTemporalTypeException(new StringBuilder().append(field.toString()).toString());
        return temporal.getLong(field)-BASE_TEMPORAL.getLong(field);
    }

    public Duration getDuration() {
        return duration;
    }

    @Override
    public String toString() {
        return dtf.format(this);
    }

    private static final DateTimeFormatter dtf = new DateTimeFormatterBuilder()
            .optionalStart()//second
            .optionalStart()//minute
            .optionalStart()//hour
            .optionalStart()//day
            .optionalStart()//month
            .optionalStart()//year
            .appendValue(ChronoField.YEAR).appendLiteral(" Years ").optionalEnd()
            .appendValue(ChronoField.MONTH_OF_YEAR).appendLiteral(" Months ").optionalEnd()
            .appendValue(ChronoField.DAY_OF_MONTH).appendLiteral(" Days ").optionalEnd()
            .appendValue(ChronoField.HOUR_OF_DAY).appendLiteral(" Hours ").optionalEnd()
            .appendValue(ChronoField.MINUTE_OF_HOUR).appendLiteral(" Minutes ").optionalEnd()
            .appendValue(ChronoField.SECOND_OF_MINUTE).appendLiteral(" Seconds").optionalEnd()
            .toFormatter();

}

查看所有这些计算,大多数单位(小时、分钟等)都有一个. tofoopart()方便方法,这可能是有帮助的。

E.g.

Duration.ofMinutes(110L).toMinutesPart() == 50

读:到父单位(小时)的下一个值的分钟数。

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。

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("")

这个答案只使用Duration方法,适用于Java 8:

public static String format(Duration d) {
    long days = d.toDays();
    d = d.minusDays(days);
    long hours = d.toHours();
    d = d.minusHours(hours);
    long minutes = d.toMinutes();
    d = d.minusMinutes(minutes);
    long seconds = d.getSeconds() ;
    return 
            (days ==  0?"":days+" days,")+ 
            (hours == 0?"":hours+" hours,")+ 
            (minutes ==  0?"":minutes+" minutes,")+ 
            (seconds == 0?"":seconds+" seconds,");
}