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,");
}
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