时间：比较两个瞬间--获得两个瞬间之间的小时数、分钟数、秒数、年数、月数[重复]

Question

public class TimePassed {
    private long seconds;
    private long minutes;
    private long hours;
    private long days;
    private long years;
    ...

    public TimePassed(double unixSeconds)  {
        Instant now = Instant.now();
        Instant ago = Instant.ofEpochSecond((long) unixSeconds);

        this.seconds = ChronoUnit.SECONDS.between(
            ago.atZone(ZoneId.systemDefault()),
            now.atZone(ZoneId.systemDefault()));  //6100
        this.minutes = ChronoUnit.MINUTES.between(
            ago.atZone(ZoneId.systemDefault()),
            now.atZone(ZoneId.systemDefault()));  //101
        this.hours = ChronoUnit.HOURS.between(
            ago.atZone(ZoneId.systemDefault()),
            now.atZone(ZoneId.systemDefault()));  //1
        this.days = ChronoUnit.DAYS.between(
            ago.atZone(ZoneId.systemDefault()),
            now.atZone(ZoneId.systemDefault()));  //0
        this.years = ChronoUnit.YEARS.between(
            ago.atZone(ZoneId.systemDefault()),
            now.atZone(ZoneId.systemDefault()));  //0
    }
}

Answer 1

基本思想，不完整：

    Duration dur = Duration.between(ago, now);

    this.seconds = dur.toSecondsPart(); // 40
    this.minutes = dur.toMinutesPart(); // 41
    this.hours = dur.toHoursPart(); // 1
    this.days = dur.toDaysPart(); // 0

与你的问题相隔6100个seoncd的瞬间测试。Java9中引入了TOXXXPart方法。对于Java8（或ThreeTen Backport)，您将需要从较粗的单元开始，即天数，并在得到下一个较细的单元之前从持续时间中减去天数。请参见lauhub的一个示例

不过，要完全正确地计算年份和天数有点困难。为了只获得超过全年的天数，这里有完整的代码：

    ZoneId zone = ZoneId.systemDefault();
    ZonedDateTime agoZdt = ago.atZone(zone);
    ZonedDateTime nowZdt = now.atZone(zone);
    this.years = ChronoUnit.YEARS.between(agoZdt, nowZdt);
    ZonedDateTime afterWholeYears = agoZdt.plusYears(this.years);

    Duration dur = Duration.between(afterWholeYears, nowZdt);

    this.seconds = dur.toSecondsPart(); // 40
    this.minutes = dur.toMinutesPart(); // 41
    this.hours = dur.toHoursPart(); // 1
    this.days = dur.toDays(); // 0